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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define stoi stoll
using pii=pair<int,int>;
#define all(c) begin(c),end(c)
#define rall(c) rbegin(c),rend(c)
#define fore(x,c) for(auto &&x:c)
#define rep(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define rrep(i, a, n) for(int i=(int)(n-1);i>=a;--i)
#define sz(c) ((int)c.size())
#define contains(c,x) (c.find(x)!=end(c))
#define inseg(l,x,r) ((l)<=(x)&&(x)<(r))
#define dump(...)
#define pb push_back
#define _ 0
const signed INF_=1001001001; const long long INF=1001001001001001001LL;
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) {
    for (auto i = begin(v); i != end(v); i++) os<<*i<<(i==end(v)-1?"":" ");return os;}
template<class T> istream& operator>>(istream &is,vector<T> &v) {
    for (auto i = begin(v); i != end(v); i++) is>>*i;return is;}
template<class T,class U> istream& operator>>(istream &is, pair<T,U> &p) {
    is>>p.first>>p.second;return is;}
template<class T, class U> bool chmax(T &a,const U &b){return a<b?a=b,1:0;}
template<class T, class U> bool chmin(T &a,const U &b){return a>b?a=b,1:0;}
template <class T> void psum(T& c) {partial_sum(begin(c), end(c), begin(c));}
template<class T> using heap=priority_queue<T,vector<T>,greater<T>>;
struct before_main_function {
    before_main_function() {
        cin.tie(0); ios::sync_with_stdio(0);
        cout << setprecision(15) << fixed;
        // #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

signed main() {
    int N, K;
    cin >> N >> K;
    string S;
    cin >> S;
    for (auto &x : S) {
        if (x == '0') x = 0;
        else x = 1;
    }

    if (count(all(S), 0) == N || count(all(S), 0) == 0) {
        cout << N << endl;
        return 0;
    }

    vector<int> l, r;
    if (S[0] == 1) {
        l.pb(0);
    } else {
        l.pb(-1);
    }
    rep(i, 1, N) {
        if (S[i - 1] != S[i] && S[i]) {
            l.pb(i);
        }
    }
    rep(i, 0, N - 1) {
        if (S[i] != S[i + 1] && !S[i]) {
            r.pb(i);
        }
    }
    if (S[N - 1] == 0) {
        r.pb(N - 1);
    }

    int L = 0, R = 0;
    while (R < sz(r) && R < K) {
        R++;
    }
    int ans = (r[R] - 1);
    if (l[L] != -1) ans -= l[L] - 1;
    while (R < sz(r)) {
        int tmp = r[R];
        if (l[L] != -1) tmp -= l[L] - 1;
        chmax(ans, tmp);
        R++;
        L++;
    }
    cout << ans << endl;
    return (0^_^0);
}