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<iostream>
#include<algorithm>
#include<set>
#include<math.h>
#include<vector>
#include<sstream>
#include<queue>
#include<functional>
#include<bitset>
#include<cstdio>
#include<iomanip>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include <string.h>

using ll = long long;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define reps(i, x) for(int i=1;i<=(int)(x);i++)
#define rrep(i, x) for(int i=((int)(x)-1);i>=0;i--)
#define rreps(i, x) for(int i=(int)(x);i>0;i--)
#define all(x) (x).begin(),(x).end()
#define m0(x) memset(x,0,sizeof(x))
#define vll vector<ll>
#define vi vector<int>
#define vpll vector<pair<ll,ll>>
#define vpi vector<pair<int,int>>
#define mod 1000000007
#define put2darr(arr, row, col) rep(i, row) { rep(j, col) { printf("%18d", arr[i][j]); } cout << endl; }

using namespace std;

template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}

/* 1..10
 * rep:   0 1 2 3 4 5 6 7 8 9
 * reps:  1 2 3 4 5 6 7 8 9 10
 * rrep:  9 8 7 6 5 4 3 2 1 0
 * rreps: 10 9 8 7 6 5 4 3 2 1
 */

int n, k;
string s;
int f(char target){
    int tmp = 0;
    int l = 0;
    int r = 1;
    while(r <= n){
        if(s[r] == target) {
            if (k)k--;
            else{
                if(tmp < r - l) tmp = r-l;
                l = r;
            }
        }else{
            if(r == n)
                if(tmp < r - l) tmp = r-l;
        }
        r++;
    }
    return tmp;
}

int main() {
    cin >> n >> k >> s;
    s = '0' + s;
    int ans =  max(f('0'), f('1'));
    cout << ans << endl;
}