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(long long int i = 0; i < n; i++)
#define _rep(i, m, n) for(long long int i = m; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int N = 1000000;
const ll mod = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

ll gcd(ll A, ll B){
    if(B == 0)return A;
    return gcd(B, A % B);
}
ll lcm(ll A, ll B){
    return A * B / gcd(A, B);
}

/*------------------------------------------------------------------*/


int main(){
    int n, k; cin >> n >> k;
    string s; cin >> s;
    int count0 = 0, count1 = 0;
    vector<int> count;
    bool flag = false; // 0 : false, 1 : true
    rep(i, n){
        if(s[i] == '0'){
            if(!flag) count0--;
            else{
                if(count1 != 0) count.push_back(count1);
                count1 = 0;
                count0 = -1;
            }
            flag = false;
        }else{
            if(flag) count1++;
            else{
                if(count0 != 0) count.push_back(count0);
                count0 = 0;
                count1 = 1;
            }
            flag = true;
        }
        if(i == n - 1){
            if(!flag) count.push_back(count0);
            else if(flag) count.push_back(count1);
        }
    }
    int l = 0;
    int m = count.size();
    rep(i, m) if(count[i] < 0) l++;
    vector<int> c_sum(m + 1);
    c_sum[0] = 0;
    _rep(i, 1, m + 1) c_sum[i] = c_sum[i - 1] + abs(count[i - 1]);
    if(l == 0 or l <= k){
        cout << c_sum[m] << endl;
        return 0;
    }
    int ans_max = 0;
    if(count[0] > 0){
        for(int i = 0; 2 * k + 1 + i <= m; i += 2){
            ans_max = max(ans_max, c_sum[2 * k + 1 + i] - c_sum[i]); 
        }
    }
    if(count[0] < 0){
        for(int i = 0; 2 * k + 1 + i <= m; i += 2){
            if(i == 0) ans_max = c_sum[2 * k];
            else{
                ans_max = max(ans_max, c_sum[2 * k + 1 + i] - c_sum[i]);
            }
        }
    }
    cout << ans_max << endl;
}