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>
using namespace std;

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

/* define const */
const long long INF = (1LL << 60);
/* end defineing */

int main()
{
    int n, k;
    deque<int> z;
    deque<int> o;
    string s;
    cin >> n >> k;
    cin >> s;

    for (int i = 0; i < n;)
    {
        if (s[i] == '0')
        {
            int count = 0;
            int j = i;
            while (s[j] == '0')
            {
                if (j >= n)
                {
                    break;
                }
                j++;
                count++;
            }
            z.push_back(count);
            i = j;
        }
        else
        {
            int count = 0;
            int j = i;
            while (s[j] == '1')
            {
                if (j >= n)
                {
                    break;
                }
                j++;
                count++;
            }
            o.push_back(count);
            i = j;
        }
    }
    /*
    for (int i = 0; i < z.size(); i++)
    {
        cout << z[i] << ",";
    }
    cout << endl;
    for (int i = 0; i < o.size(); i++)
    {
        cout << o[i] << ",";
    }
    cout << endl;
    */
    int tmp_max = -1;
    int i = 0;

    if (s[0] == '0')
    {
        int tmp = 0;
        int j;
        for (j = 0; j < z.size() && j < k; j++)
        {
            tmp += z[j];
        }
        j--;
        int cc = 0;
        for (int m = j; m >= 0; m--)
        {
            if (cc > k)
            {
                break;
            }
            tmp += o[m];
            cc++;
        }
        tmp_max = tmp;
        // cout << tmp_max << endl;

        for (int n = k; n < z.size(); n++)
        {
            tmp -= z[n - k];
            tmp += z[k];
            if (n - k - 1 >= 0)
            {
                tmp -= o[n - k - 1];
            }
            tmp += o[k];
            if (tmp_max < tmp)
                tmp_max = tmp;

            // cout << tmp << endl;
        }
    }
    else
    {
        int tmp = 0;
        int j;
        for (j = 0; j < z.size() && j < k; j++)
        {
            tmp += z[j];
        }
        j--;
        int cc = 0;
        for (int m = j + 1; m >= 0; m--)
        {
            if (cc > k)
            {
                break;
            }
            tmp += o[m];
            cc++;
        }
        tmp_max = tmp;
        // cout << tmp << endl;

        for (int n = k; n < z.size(); n++)
        {
            tmp -= z[n - k];
            tmp += z[k];
            if (n - k - 1 >= 0)
            {
                tmp -= o[n - k - 1];
            }
            tmp += o[k];
            if (tmp_max < tmp)
                tmp_max = tmp;

            // cout << tmp << endl;
        }
    }
    cout << tmp_max << endl;
}