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 <string> 
#include <vector>
#include <iomanip>
using namespace std;
#define lol long long


int main()
{
	int N, K;
	string S;
	cin >> N >> K >> S;
	vector<int> R(N+1);
	vector<int> Rui(N + 1);
	int ichi = (N + 2) / 2;
	int t = 0;
	if (S[0] == '0') {
		t = 1;
		R[1]++;
	}
	else {
		R[0]++;
	}
	for (int i = 0; i < N - 1; i++) {
		if (S[i] == S[i + 1]) {
			R[t]++;
		}
		else {
			t++;
			R[t]++;
		}
	}
	Rui[0] = R[0];
	for (int i = 1; i < N + 1; i++) {
		Rui[i] = Rui[i - 1] + R[i];

	}
	int ans = 0;
	for (int i = 0; i < ichi - K; i++) {
		ans = max(ans, Rui[i * 2 + K * 2] - Rui[i * 2]);
	}
	int bun = 0;
	if (S[0] == '1' || S[N - 1] == '1') {
		bun++;
	}
	cout << ans + bun << endl;
}