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(arg) {
    const N = arg.split("\n")[0].split(" ").map((x)=>{return parseInt(x)})[0]
    const K = arg.split("\n")[0].split(" ").map((x)=>{return parseInt(x)})[1]
    const S = arg.split("\n")[1].split("").map((x)=>{return parseInt(x)})
    let SS = [0], i, tmp = 1, ans = 0
    for (i = 0; i < N; ++i){
        if (S[i] == tmp){
            SS[SS.length-1]++
        }else{
            SS.push(1)
        }
        tmp = S[i]
    }
    if(SS.length%2 === 0) SS.push(0)
    for(i = 0; i < SS.length - 2*K; i+=2){
        tmp = SS.slice(i, i+2*K+1).reduce((a,x) => a+=x,0)
        if (tmp > ans) ans = tmp
    }
    console.log(ans)
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));