Codeforces round #272 (Div. 2)

Link: http://codeforces.com/contest/476

D Question YY, ABC water

A. dreamoon and stairstime limit per test1 secondmemory limit per test256 megabytes

Dreamoon wants to climb up a stairNSteps. He can climb 1 or 2 steps at each move. dreamoon wants the number of moves to be a multiple of an integerM.

What is the minimal number of steps making him climb to the top of the stairs that satisfies his condition?

Input

The single line contains two space separated IntegersN,M(0? <?N? ≤? 10000 ,? 1? <?M? ≤? 10 ).

Output

Print a single integer-the minimal number of moves being a multipleM. If there is no way he can climb satisfying condition print? -? 1 instead.

Sample test (s) Input

10 2

Output

6

Input

3 5

Output

-1

Note

For the first sample, dreamoon cocould climb in 6 moves with following sequence of steps: {2, 2, 2, 1, 1 }.

For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1} with 2, 2, and 3 steps respectively. all these numbers are not multiples of 5.

<span style="font-size:14px;">#include <cstdio>int main(){    int n, m;    scanf("%d %d", &n, &m);    int temp = n / 2;    if(n % 2)         temp++;    if(temp % m == 0)         printf("%d\n", temp);    else    {        temp = temp + m - temp % m;        if(temp > n)             printf("-1\n");        else             printf("%d\n", temp);    }}</span>

B. dreamoon and wifitime limit per test1 secondmemory limit per test256 megabytes

Dreamoon is standing at the position 0 on a number line. drazil is sending a list of commands through Wi-Fi to dreamoon's smartphone and dreamoon follows them.

Each Command is one of the following two types:

1. Go 1 unit towards the positive ction, denoted as '+'
2. Go 1 unit towards the negative ction, denoted '-'

But the Wi-Fi condition is so poor that dreamoon's smartphone reports some of the commands can't be recognized and dreamoon knows that some of them might even be wrong though successfully recognized. dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the1 unit to the negative or positive direction with the same probability0.5 ).

You are given an original list of commands sent by drazil and list stored ed by dreamoon. What is the probability that dreamoon ends in the position originally supposed to be final by drazil's commands?

Input

The first line contains a stringS_{1-The commands drazil sends to dreamoon, this string consists of only the characters in the set {'+ ','-'}.}

The second line contains a stringS_{2-The commands dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+ ','-','? '}.'? 'Denotes an unrecognized command.}

Lengths of two strings are equal and do not exceed10.

Output

Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn' t exceed10^{? -? 9.}

Sample test (s) Input

++-+-+-+-+

Output

1.000000000000

Input

+-+-+-??

Output

0.500000000000

Input

+++??-

Output

0.000000000000

Note

For the first sample, bothS_{1 andS2 will lead dreamoon to finish at the same position? +? 1.}

For the second sample,S_{1 will lead dreamoon to finish at position 0, while there are four possibilitesS2: {"+-++", "+-", "+ -- +", "+ ---"} with ending position {+ 2, 0, 0, -2} respectively. so there are2 correct cases out of 4, so the probability of finishing at the correct position is0.5.}

For the third sample,S_{2 cocould only lead us to finish at positions {+ 1,-1,-3}, so the probability to finish at the correct position? +? 3 is 0.}

<span style="font-size:14px;">#include <cmath>#include <cstdio>#include <cstring>char s1[15],s2[15];int main(){    scanf("%s %s",s1,s2);    int p1=0, m1=0;    int p2=0, m2=0;    int size=(int)strlen(s1);    int wh=0;    for(int i=0;i<size; i++)    {        if(s1[i]=='+') ++p1;        else ++m1;    }    for(int i=0;i<size; i++)    {        if(s2[i]=='+') ++p2;        else if(s2[i]=='-') ++m2;        else ++wh;    }    if(wh == 0)    {        if(p1 - m1 == p2 - m2)            printf("1\n");        else            printf("0\n");        return 0;    }    int all = 1;    for(int i = 0;i < wh; i++)        all *= 2;    int pneed,mneed;    int temp = p1 - m1 - p2 + m2;    if(fabs(temp) > wh)    {        printf("0\n");        return 0;    }    pneed = (temp + wh) / 2;    mneed = wh - pneed;    int ans = 1;    for(int i = 0; i < pneed; i++)    {        ans *= wh;        --wh;    }    for(int i=1; i<= pneed; i++)        ans /= i;    printf("%.12f\n",double(ans)/double(all));}</span>

C. dreamoon and sumstime limit per test1.5 secondsmemory limit per test256 megabytes

Dreamoon loves summing up something for no reason. One day he obtains two integersAAndBOccasionally. He wants to calculate the sum of allnice integers. Positive IntegerXIs called nice if and, whereKIs some integer number in range [1 ,?A].

By we denote thequotient of integer divisionXAndY. By we denote theremainder of integer divisionXAndY. You can read more about these operations here: http://goo.gl/AcsXhT.

The answer may be large, so please print its remainder modulo1? 000? 000? 007 (10^{9? +? 7). Can you compute it faster than dreamoon?}

Input

The single line of the input contains two integersA,B(1? ≤?A,?B? ≤? 10^{7 ).}

Output

Print a single integer representing the answer modulo1? 000? 000? 007 (10^{9? +? 7 ).}

Sample test (s) Input

1 1

Output

0

Input

2 2

Output

8

Note

For the first sample, there are no nice integers because is always zero.

For the second sample, the set of Nice integers is {3 ,? 5 }.

<span style="font-size:14px;">#include <cstdio>#define ll long longconst int mod=1e9 + 7;int main(){    ll a,b;    scanf("%I64d %I64d", &a, &b);    if(b == 1)         printf("0\n");    else    {        ll res = (b * (b - 1) / 2) % mod;        ll sum = (a * (a + 1) / 2) % mod;        ll ans = (a % mod * res % mod) % mod;        ll temp = (sum % mod * res % mod) % mod;        printf("%I64d\n", (ans % mod + (temp % mod * b % mod)) % mod);    }}</span>

D. dreamoon and setstime limit per test1 secondmemory limit per test256 megabytes

Dreamoon likes to play with sets, integers and. is defined as the largest positive integer that divides bothAAndB.

LetSBe a set of exactly four distinct integers greater than0.defineSTo be of rankKIf and only if for all pairs of distinct elementsS_I,SJFromS,.

GivenKAndN, Dreamoon wants to make upNSets of rankKUsing integers from1MSuch that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimumMThat makes it possible and print one possible solution.

Input

The single line of the input contains two space separated IntegersN,K(1? ≤?N? ≤? 10? 000 ,? 1? ≤?K? ≤? 100 ).

Output

On the first line print a single integer-the minimal possibleM.

On each of the nextNLines Print four space separated integers representingI-Th set.

Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimalM, Print any one of them.

Sample test (s) Input

1 1

Output

51 2 3 5

Input

2 2

Output

222 4 6 2214 18 10 16

Note

For the first example it's easy to see that set {1 ,? 2 ,? 3 ,? 4} isn' t a valid set of rank 1 since.

<span style="font-size:14px;">#include <cstdio>int main(){    int n, k;    scanf("%d %d", &n, &k);    int a = 1, b = 2, c = 3, d = 5;    printf("%d\n", (d * k + 6 * k * (n- 1)));    a *= k; b *= k; c *= k; d *= k;    for(int i = 0; i < n; i++)    {        printf("%d %d %d %d\n",a, b, c, d);        a += 6 * k;        b += 6 * k;        c += 6 * k;        d += 6 * k;    }}</span>

Codeforces round #272 (Div. 2)