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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
typedef long long ll;
using namespace std;
struct D {
    
    int left;
    int right;
    int attr;
};
int main() {
    int n, k; cin >> n >> k;
    string s; cin >> s;
    vector<D> vc;
    int left = 0;
    int target = s[0] - '0';
    //ランレングス
    for(int i=0;i<n;i++) {
        if(i==n-1) {
            int target = s[i] - '0';
            vc.push_back({left, i, target});
        }
        else if(s[i] - '0' == s[i+1] - '0') {
        }else{
            int target = s[i] - '0';
            vc.push_back({left, i, target});
            left = i+1;
        }
    }
    /*rep(i, vc.size()) {
        printf("%d %d %d\n",vc[i].left,vc[i].right,vc[i].attr);
    }*/
    vector<D> vc0;
    rep(i, vc.size()) {
        if(vc[i].attr == 0) {
            D x = {vc[i].left,vc[i].right,vc[i].attr};
            if(i!=0)
                x.left = vc[i-1].left;
            if(i!=vc.size()-1)
                x.right = vc[i+1].right;
            vc0.push_back(x);
        }
    }
    int mx = 0;
    int sz = vc0.size();
    if(sz == 0) mx = n;
    rep(i,sz) {
        if(i+k-1>=sz) break;
        mx = max(mx,vc0[i+k-1].right - vc0[i].left + 1);
    }
    cout << mx;
}