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: Accepted)

function sum_array(arr){
  var sum=0;
  var i=0
  while(i<arr.length){
    sum = sum + arr[i];
    i=i+1
  }
  return sum;
}

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) );

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

  var compressed_numbers = [];
  if(input_numbers[0]==0){compressed_numbers=[0];}

  var pre_count = 1;
  var pre_number=input_numbers[0];

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

  var accumulation_sum_array = [0];
  for(var i=0;i<compressed_numbers.length;i++){
    accumulation_sum_array[i+1] = accumulation_sum_array[i] + compressed_numbers[i]
  }

//console.log(compressed_numbers);
//console.log(accumulation_sum_array);

  var answer=0;
  var sub_array=[];
  const a_a_length = accumulation_sum_array.length;
  const fetch_count = k*2+1;
  for(var  i=0 ; i<compressed_numbers.length ; i+=2){
    // k * 2 + 1 要素をとってくる
    if( i+fetch_count > a_a_length-1 ){
      answer = Math.max(accumulation_sum_array[a_a_length-1] - accumulation_sum_array[i], answer)
    } else {
      answer = Math.max(accumulation_sum_array[i+fetch_count] - accumulation_sum_array[i], answer)
    }
  }

  console.log(answer);
}

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