Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define REP(i,N) for(i=0;i<N;i++)
typedef long long ll;

int main(void){
  int i,index=0,count=0,N,K;
  string S;
  cin >> N >> K;
  cin >> S;
  vector<int> comp;
  vector<int> len(1,0);
  comp.push_back(0);
  char pre=' ';
  REP(i,N){
    if((i!=0 && pre!=S[i])){
      if(S[i]=='1')comp.push_back(-count);
      else comp.push_back(count);
      count=0;
    }
    pre = S[i];
    count++;
  }
  if(S[N-1]=='1')comp.push_back(count);
  else comp.push_back(-count);

  count = 0;
  REP(i,comp.size()){
    if(comp[i]<0){
      if(K<=count){
        len.push_back(len[index]+comp[i-2*K]-comp[i-2*K-1]);
        index++;
      }
      len[index]-=comp[i];
      count++;
    }else{
      len[index]+=comp[i];
    }
  }
  int max=0;
  REP(i,len.size()){
    if(max<len[i])max=len[i];
  }
  cout << max << endl;
  return 0;
}