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 Main(input){
  input = input.split("\n");
  inputStr = input[0].split(" ");
  //入力値NとKを数値化して確保
  var N = Number.parseInt(inputStr[0]);
  var K = Number.parseInt(inputStr[1]);
  //入力値を配列に入れる
  var S = input[1];
  //入力値S_arrを、0と1の塊別に纏めて、配列A_arrに入れる。0はマイナス化する。
  var A_arr = [];
  var cnt = 1;
  var flg = 0;
  for( var i=0; i < N-1; i++){
    if( S.substr(i,1) !== S.substr(i+1,1) ){
      if(S.substr(i,1)==="0"){
        A_arr.push(cnt * -1);
        flg = 1;
      } else {
        A_arr.push(cnt);
      }
      cnt = 1;
    } else {
      cnt += 1;
    }
  }
  if(S.substr(i,1)==="0"){
    A_arr.push(cnt * -1);
    flg = 1;
  } else {
    A_arr.push(cnt);
  }
  
  //console.log("配列",A_arr);
  
  //ループカウンタiとjを定義し、iとjの間にマイナス値が5つ含んだ状態での尺取法を行う。
  var m_cnt = 0;
  var tmp = 0;
  var max = 0;
  var j=0;
  var lsat_j = -1;
  //console.log("配列の長さ",A_arr.length);
  for( var i=0 ; i < A_arr.length ; i++){
    //範囲の中にマイナス値をK個含むか、配列終端に辿り着くまでループしてtmpに足し込む
    while( m_cnt < K && j < A_arr.length ){
      if(lsat_j !== j ){
        tmp += Math.abs(A_arr[j]);
        //console.log("tmpに値を足しこみ",A_arr[j]);
        lsat_j = j;
      }
      if( A_arr[j] < 0 ){
        m_cnt +=1;
      }
      j++;
    }
    if(lsat_j !== j ){
      lsat_j = j;
      if( m_cnt >= K && j < A_arr.length ){
        tmp += Math.abs(A_arr[j]);
        //console.log("tmpに値を足しこみ",A_arr[j]);
      }
    }
    //最大値比較
    //console.log("iとj",i,j);
    //console.log("maxとtmp",max,tmp);
    max = Math.max(max,tmp);
    //マイナス値の消しこみ
    if( A_arr[i] < 0 ){
      m_cnt -= 1;
    }
    tmp -= Math.abs(A_arr[i]);
    //console.log("tmpから値を消しこみ",A_arr[i]);
  }
  console.log(max);
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));