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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using static System.Math;

class Solver {
    Scanner sc = new Scanner();

    public void Solve() {
        var N = sc.nextInt();
        var K = sc.nextInt();
        var S = sc.next();
        var lst = new List<int>();
        var cnt = 1;
        for (int i = 1; i < N; i++) {
            if (S[i] == S[i-1]) {
                cnt++;
            } else {
                lst.Add(cnt);
                cnt = 1;
            }
        }
        lst.Add(cnt);

        var len = 0;
        if (S[0] == '1') {
            len = (2 * K) + 1;
        } else {
            len = 2 * K;
        }

        if (len >= lst.Count) {
            WriteLine(lst.Sum());
            return;
        }

        var sum = lst.Take(len).Sum();
        var ans = sum;
        len = (2 * K) + 1;
        var first = 0;
        if (S[0] == '0') {
            first = 1;
        }
        sum = lst.Skip(first).Take(len).Sum();
        ans = Max(ans, sum);


        for (int i = first + len + 1; i < lst.Count; i += 2) {
            sum += lst[i] + lst[i - 1];
            sum -= lst[i - len] + lst[i - (len + 1)];
            ans = Max(ans, sum);
        }

        if (S[N-1] == '1') {
            len = (2 * K) + 1;
        } else {
            len = 2 * K;
        }
        lst.Reverse();
        sum = lst.Take(len).Sum();
        ans = Max(ans, sum);


        WriteLine(ans);
    }
}

class Scanner {
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner() {
        s = new string[0];
        i = 0;
    }

    public string next() {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt() {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0) {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong() {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0) {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble() {
        return double.Parse(next());
    }

    public double[] ArrayDouble(int N, double add = 0) {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}

class Program {
    static void Main(string[] args) {
        Solver s = new Solver();
        s.Solve();
    }
}