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){
  input=input.trim().split(/\s+/);
  var N=input.shift()-0;
  var K=input.shift()-0;
  input=input.join('');
  var S=input.split('').map(x=>x-0);
  var c=[1];
  for(let i=0;i<N-1;i++){
    if(S[i+1]===S[i]){
      c[c.length-1]++;
    }else{
      c.push(1);
    }
  }
  if(S[0]===0)c.unshift(0);
  if(c.length%2===0)c.push(0);
  var L=2*K+1;
  var v=[0];
  for(let j=0;j<L;j++){
    v[0]+=c[j];
  }
  for(let j=0;j<(c.length-L)/2;j++){
    v[j+1]=v[j]+c[2*j+L]+c[2*j+L+1]-c[2*j]-c[2*j+1];
  }
  var result=Math.max.apply(null,v);
  console.log(result);
}

Main(require('fs').readFileSync('/dev/stdin','utf8'));