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>
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define RFOR(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define rep(i,n) FOR(i,0,(n))
#define REPS(i,n) FOR(i,1,(n)+1)
#define RREP(i,n) RFOR(i,0,(n))
#define rng(x) (x).begin(),(x).end()
#define rrng(x) (x).rbegin(),(x).rend()
#define debug(x) cout << #x << " = " << (x) << endl;
#define SORT(x) sort(rng(x))
#define RSORT(x) sort(rrng(x))
#define SUM(x) accumulate(rng(x),0)
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define uni(x) x.erase(unique(rng(x)),x.end())
#define SP << " " <<
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using pii = pair<int, int>;
using tiii = tuple<int, int, int>;
const int mod = 1000000007;
const double EPS = 1e-9;
const int INF = (1<<30)-1;
const ll INFLL = (1LL<<62)-1;
const int dx[4] = {1,0,-1,0};
const int dy[4] = {0,1,0,-1};
//cout << fixed << setprecision(10);

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}

int main () {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;

  vi nums;
  int now = 1;
  int cnt = 0;
  // 1-0-1-0-1の形の配列を作る
  // 0番目の区間が0なら、(1,0)が先頭に追加される
  // 0番目の区間が1なら、そのまま数えられる
  rep(i,n) {
    if (s[i] == (char)('0' + now)) cnt++;
    else {
      nums.pb(cnt);
      now = 1 - now;
      cnt = 1;
    }
  }
  if (cnt != 1) nums.pb(cnt);
  if (nums.size()%2 == 0) nums.pb(0);

  int add = 2*k + 1;
  int ans = 0;
  for (int i = 0; i < nums.size(); i+=2) {
    int tmp = 0;

    int left = i;
    int right = min(i+add, (int)nums.size());
    for (int j = left; j < right; ++j) {
      tmp += nums[j];
    }
    ans = max(tmp, ans);
  }
  cout << ans << endl;
  return 0;
}