Codeforces Round # Pi (Div. 2) D. One-dimen1_battle Ships binary stl Application

Codeforces Round # Pi (Div. 2) D. One-dimen1_battle Ships binary stl Application

 

D. One-dimen1_battle Ships time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Alice and Bob love playing one-dimen1_battle ships. They play on the field in the form of a line consistingNSquare cells (that is, on a 1 hour × hourNTable ).

At the beginning of the game Alice putsKShips on the field without telling their positions to Bob. Each ship looks as a 1 minute × hourARectangle (that is, it occupies a sequenceAConsecutive squares of the field). The ships cannot intersect and even touch each other.

After that Bob makes a sequence of shots. He names cells of the field and Alice either says that the cell is empty (miss), or that the cell belongs to some ship (hit ).

But here's the problem! Alice like to cheat. May be that is why she responds to each Bob's move with a miss.

Help Bob catch Alice cheating-find Bob's first move, such that after it you can be sure that Alice cheated.

Input

The first line of the input contains three integers:N,KAndA(1 digit ≤ DigitN, Bytes,K, Bytes,ALimit ≤ limit 2. 105)-the size of the field, the number of the ships and the size of each ship. It is guaranteed thatN,KAndAAre such that you can putKShips of sizeAOn the field, so that no two ships intersect or touch each other.

The second line contains integerM(1 digit ≤ DigitMLimit ≤ limitN)-The number of Bob's moves.

The third line containsMDistinct integersX1, bytes,X2, middle..., middle ,...,XM, WhereXIIs the number of the cell where Bob madeI-Th shot. The cells are numbered from left to right from 1N.

Output

Print a single integer-the number of such Bob's first move, after which you can be sure that Alice lied. Bob's moves are numbered from1MIn the order the were made. If the sought move doesn't exist, then print-1.

Sample test (s) input

11 3 354 8 6 1 11

Output

3

Input

5 1 321 5

Output

-1

Input

5 1 313

Output

1

The meaning of the question, given a range, removing a vertex each time, requires that you put down the remaining blank space. k intervals are a-long, and you can simply use set to save the current split interval. The total complexity is n * logn.

#define N 200050#define M 200050#define maxn 205#define MOD 1000000000000000007#define mem(a)  memset(a, 0, sizeof(a))int k,len,m,n,pri[N],num;set
 
   myset;set
  
   ::iterator it;int main(){    while(S(n)!=EOF)    {        S2(k,len);        S(m);        pii tp = mp(1,n);        myset.insert(tp);        int ans = -1;        num = (tp.second - tp.first + 2) / (len + 1);        bool flag = true;        FI(m){            scan_d(pri[i]);            if(flag){                it = myset.lower_bound(mp(pri[i],n + 1));                it --;                pii p = *it;                {                    myset.erase(it);                    num -= (p.second - p.first + 2) / (len + 1);                    if(p.first  <= pri[i] - 1){                        pii tp = mp(p.first,pri[i] - 1);                        num += (tp.second - tp.first + 2) / (len + 1);                        myset.insert(tp);                    }                    if(pri[i] + 1 <= p.second){                        pii tp = mp(pri[i] + 1,p.second);                        num += (tp.second - tp.first + 2) / (len + 1);                        myset.insert(tp);                    }                    if(num < k){                        ans = i + 1;                        flag = false;                    }                }            }        }        printf(%d,ans);    }    return 0;}