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: Time Limit Exceeded)

function sum_array(arr){
  var sum=0;
  for(var i=0 ; i<arr.length ; i++ ){
    sum += arr[i];
  }
  return sum;
}

function max_array(arr){
  var max_num=0;
  for(var max_index = 0 ; max_index<arr.length ; max_index++){
    max_num = Math.max(max_num, arr[max_index])
  }
  return max_num;
}

function main(input) {
  const inputs = input.split('\n');
  const counts = inputs[0].split(' ').map( x => parseInt(x, 10) );

  const n = counts[0];
  const k = counts[1];

  const input_numbers = inputs[1].split('').map( x => parseInt(x, 10) );
  const first_element = input_numbers[0];

if(n==1){
  console.log(1);
  return;
}

  compressed_numbers = [];
  pre_numbers = [input_numbers[0]];
  for(var  i=1 ; i<n ; i++){
    if(pre_numbers[0] != input_numbers[i]){
      compressed_numbers.push(pre_numbers.length);
      pre_numbers=[input_numbers[i]];
    }else{
      pre_numbers.push(input_numbers[i]);
    }
  }
  if(pre_numbers.length>0){
    compressed_numbers.push(pre_numbers.length);
  }

  answer=0;
  sub_array=[];
  for(var  i=0 ; i<n ; i++){
    if( i%2==0 ){
      if(first_element==0){
        // k * 2 要素をとってくる
        num = (i+k*2)
        sub_array = compressed_numbers.slice(i, num )
        answer = Math.max(answer, sum_array(sub_array));
      } else {
        // k * 2 + 1 要素をとってくる
        num = (i+k*2+1)
        sub_array = compressed_numbers.slice(i, num )
        answer = Math.max(answer, sum_array(sub_array));
      }
    } else {
      if(first_element==0){
        // k * 2 + 1 要素をとってくる
        num = (i+k*2+1)
        sub_array = compressed_numbers.slice(i, num )
        answer = Math.max(answer, sum_array(sub_array));
      } else {
        // k * 2 要素をとってくる
        num = (i+k*2)
        sub_array = compressed_numbers.slice(i, num )
        answer = Math.max(answer, sum_array(sub_array));
      }
    }
  }

  console.log(answer);
}

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