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.split("\n");
  var N = parseInt(input[0].split(" ")[0], 10);
  var K = parseInt(input[0].split(" ")[1], 10);
  var S = [...input[1]].map(function (v) {
    return parseInt(v, 10);
  });
  var array = [];
  var now = 1;
  var localLen = 0;

  for (var i = 0; i < N; i++) {
    if (S[i] === now) {
      localLen++;
    } else {
      array.push(localLen);
      now = S[i];
      localLen = 1;
    }
  }

  array.push(localLen);

  if (array.length % 2 == 0) {
    array.push(0);
  } //console.log(array);


  var sum = [];
  var final = array.reduce(function (a, c) {
    sum.push(a);
    return a + c;
  });
  sum.push(final); //console.log(sum);

  var max = sum[0];

  for (var _i = 0; _i < sum.length - 2 * K; _i += 2) {
    var right = _i + 2 * K;
    var left = _i - 1; //console.log(left, right);

    var lSum = sum[right];
    lSum -= left >= 0 ? sum[left] : 0;

    if (lSum > max) {
      max = lSum;
    }
  }

  console.log(max);
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));