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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

"use strict";
function Main(input) {
    let lines = input.split("\n");
    let nk = lines.shift().split(/\s/).map((v) => v - 0);
    let N = nk[0], K = nk[1];
    let S = lines.shift().split(/\s/)[0].split("").map((v) => v - 0);
    let n = 0, Z = [];
    let p, b = S[0], i = 0;
    while (n < N) {
        for (p = n; p < N && S[p] == b; p++) { }
        let l = p - n;
        Z.push({kind: b, pos: n, len: l, index: i++});
        b = (b + 1) % 2;
        n = p;
    }
    let L = Z.length, LL = L - (K - 1) * 2;
    let start = Z[0].kind;
    let ans = 0;
    for (let n = 0; n < LL; n += 2) {
        let sum = 0, m, k;
        for (m = n, k = 0; m < L && k < K; k++, m += 2) {
            if (m < L) sum += Z[m].len;
            if (m + 1 < L) sum += Z[m + 1].len;
        }
        if (start == 1 && m < L) sum += Z[m].len;
        if (ans < sum) ans = sum;
    }
    console.log(ans);
}
if (typeof window == "undefined") Main(require("fs").readFileSync("/dev/stdin", "utf8"));