Codeforces round #196 (Div. 2) C. Quiz

C. quiztime limit per test

1 second

Memory limit per test

256 megabytes

Input

Standard Input

Output

Standard output

Manao is taking part in a quiz. The quiz consistsNConsecutive questions. A Correct answer gives one point to the player. The game also has a counter
Of consecutive correct answers. when the player answers a question correctly, the number on this counter increases by 1. if the player answers a question incorrectly, the counter is reset, that is, the number on it has CES to 0. if after an answer the counter
Reaches the numberK, Then it is reset, and the player's score is doubled. Note that in this case, first 1 point is added to the player's score, and then
The total score is doubled. At the beginning of the game, both the player's score and the counter of consecutive correct answers are set to zero.

Manao remembers that he has answered exactlyMQuestions Correctly. But he does not remember the order in which the questions came. He's trying to figure
Out what his minimum score may be. Help him and compute the remainder of the corresponding number after division by 1000000009 (109 bytes + limit 9 ).

Input

The single line contains three space-separated IntegersN,MAndK(2 cores ≤ CoresKLimit ≤ limitNMemory ≤ memory 109; 0 memory ≤ memoryMLimit ≤ limitN).

Output

Print a single integer-the remainder from division of manao's minimum possible score in the quiz by 1000000009 (109 bytes + limit 9 ).

Sample test (s) Input

5 3 2

Output

3

Input

5 4 2

Output

6

Note

Sample 1. manao answered 3 questions out of 5, and his score wocould double for each two consecutive correct answers. if manao had answered the first, third and th questions, he wowould have scored as much as 3 points.

Sample 2. Now manao answered 4 questions. The minimum possible score is obtained when the only wrong answer is to the question 4.

Also note that you are asked to minimize the score and not the remainder of the score modulo 1000000009. For example, if manao cocould obtain either 2000000000 or 2000000020 points,
The answer is 2000000000MoD1000000009, even though2000000020MoD1000000009 is
A smaller number.

To use matrix multiplication, or else it will time out, we can enumerate from the back to the front, mark every K, as wrong, then we can introduce the formula, f [N] = (F [n-1] * 2 + 2, so that we can use matrix multiplication to quickly calculate the answer!

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define mod 1000000009struct matrix {    __int64 a[2][2];    matrix operator * (matrix b)const    {        matrix temp;        int i,j,k;        for(i=0;i<2;i++)        for(j=0;j<2;j++)        {            temp.a[i][j]=0;            for(k=0;k<2;k++)            {                temp.a[i][j]+=a[i][k]*b.a[k][j];                temp.a[i][j]%=mod;            }        }        return temp;    }};__int64 k;__int64 fcos(__int64 n){    if(n<=0)    return 0;    matrix start,temp,end;    end.a[0][0]=0,end.a[0][1]=2,end.a[1][0]=end.a[1][1]=0;    start.a[0][0]=1,start.a[0][1]=0,start.a[1][0]=0,start.a[1][1]=1;    temp.a[0][0]=2,temp.a[0][1]=0,temp.a[1][1]=temp.a[1][0]=1;    while(n)    {        if(n&1)        {            start=start*temp;        }        temp=temp*temp;        n=n>>1;    }    end=end*start;    return end.a[0][0]*k%mod;}__int64 fmin(int a,int b){    if(a<b)    return a;    return b;}int main(){    __int64 n,m,no;    //printf("%I64d",fcos(2));    while(scanf("%I64d%I64d%I64d",&n,&m,&k)!=EOF)    {        int mn=n-m;        m=fmin(n-m,n/k);        no=(n-m*k)/k;        printf("%I64d\n",(n-no*k-mn+fcos(no))%mod);    }    return 0;}