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: Wrong Answer)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
ll MOD = 1e9+7;

#define vec vector<int>
#define vecll vector<ll>
#define vpair vector<pair<int, int>>
#define vpairll vector<pair<ll, ll>>

#define rep(i,n)         for(int i=(int)0; i<(int)n; i++)
#define REP(i,m,n)       for(int i=(int)m; i<(int)n; i++)
#define arr(var, n)      vec var(n); rep(i,n){cin >> var[i];}
#define arrll(var, n)    vecll var(n); rep(i,n){cin >> var[i];}
#define all(var)         (var).begin(), (var).end()
#define sortall(var)     sort(all(var))
#define prt(var)         cout << var << "\n"
#define prtd(n, var)     cout << fixed << setprecision(n) << var << "\n"
#define prt2(v1, v2)     cout << v1 << " " << v2 << "\n"
#define prt3(v1, v2, v3) cout << v1 << " " << v2 << " " << v3 << "\n"
#define prtall(var)      rep(i,var.size()){cout << var[i] << " ";} cout << "\n"

//----------------------------------------------------------------

int main(void) {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;

  vec a;
  bool is_head1=false;
  if(s[0]=='1'){
    is_head1=true;
  }

  char tmp='s';
  rep(i,n){
    if(s[i]!=tmp){
      a.push_back(i);
      tmp=s[i];
    }
  }
  //prt("print a");
  //prtall(a);
  int ans=0;
  if((s[0]=='1' && a.size()<=2*k+1) || (s[0]=='0' && a.size()<=2*k)){
    ans=n; prt(ans); return 0;
  }

  a.push_back(n);
  int count=0;
  rep(i,a.size()){
    if(s[a[i]]=='0'){
      if(i+2*k+1>a.size()){
        break;
      }
      count=a[i+2*k+1]-a[i];
      ans=max(ans, count);
    } else {
      if(i+2*k>a.size()){
        break;
      }
      count=a[i+2*k]-a[i];
      ans=max(ans, count);
    }
  }
  //prt("===ANS===");
  prt(ans);

}