#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pi = pair<ll, ll>;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll minf = 0xc0c0c0c0c0c0c0c0;
const ll mod = 998244353;
const ll MAXN = 5050;
ll dp[5050];
ll mx[5050];
ll fact[MAXN] = {1, 1}, inv[MAXN] = {1, 1}, fact_inv[MAXN] = {1, 1};
void init() {
  for (ll i = 2; i < MAXN; i++) {
    fact[i] = ll(fact[i - 1]) * i % mod;
  }
  for (ll i = 2; i < MAXN; i++) {
    inv[i] = ll(inv[mod % i]) * (mod - mod / i) % mod;
  }
  for (ll i = 2; i < MAXN; i++) {
    fact_inv[i] = ll(fact_inv[i - 1]) * inv[i] % mod;
  }
}
ll nCr(ll n, ll r) {
  return n >= r && r >= 0
             ? ll(fact[n]) * fact_inv[n - r] % mod * fact_inv[r] % mod
             : 0;
}
ll nPr(ll n, ll r) {
  return n >= r && r >= 0 ? ll(fact[n]) * fact_inv[n - r] % mod : 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(20);
  init();
  ll n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 0) {
    cout << 1 << '\n';
    return 0;
  }
  memset(mx, -1, sizeof(mx));
  for (ll l = 0, r = -1, c = 0;;) {
    while (r < n - 1 && c + (s[r + 1] == '1') <= k) {
      ++r;
      c += s[r] == '1';
    }
    if (c < k) {
      break;
    }
    mx[l] = r;
    c -= s[l] == '1';
    ++l;
  }
  for (ll i = n - 1; i >= 0; i--) {
    if (mx[i] == -1) {
      dp[i] = 1;
    } else if (mx[i] == n - 1) {
      dp[i] = nCr(n - i, k);
    } else {
      dp[i] = dp[i + 1];
      ll R = mx[i];
      if (s[i] == '0') {
        dp[i] += nCr(R - i, k - 1);
        if (dp[i] >= mod) {
          dp[i] -= mod;
        }
      } else {
        dp[i] += nCr(R - i, k);
        if (dp[i] >= mod) {
          dp[i] -= mod;
        }
      }
    }
  }
  cout << dp[0] << '\n';
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

You are given a binary string (i. e. a string consisting of characters 0 and/or 1) s of length n. You can perform the following operation with the string s at most once: choose a substring (a contiguous subsequence) of s having exactly k characters 1 in it, and shuffle it (reorder the characters in the substring as you wish).

Calculate the number of different strings which can be obtained from s by performing this operation at most once.

Input

The first line contains two integers n and k (2 ≤ n ≤ 5000; 0 ≤ k ≤ n).

The second line contains the string s of length n, consisting of characters 0 and/or 1.

Output

Print one integer — the number of different strings which can be obtained from s by performing the described operation at most once. Since the answer can be large, output it modulo 998244353.

Examples

Input


7 2
1100110


Output


16


Input


5 0
10010


Output


1


Input


8 1
10001000


Output


10


Input


10 8
0010011000


Output


1

Note

Some strings you can obtain in the first example:

  * to obtain 0110110, you can take the substring from the 1-st character to the 4-th character, which is 1100, and reorder its characters to get 0110; 
  * to obtain 1111000, you can take the substring from the 3-rd character to the 7-th character, which is 00110, and reorder its characters to get 11000; 
  * to obtain 1100101, you can take the substring from the 5-th character to the 7-th character, which is 110, and reorder its characters to get 101. 



In the second example, k = 0 so you can only choose the substrings consisting only of 0 characters. Reordering them doesn't change the string at all, so the only string you can obtain is 10010.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pi = pair<ll, ll>;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll minf = 0xc0c0c0c0c0c0c0c0;
const ll mod = 998244353;
const ll MAXN = 5050;
ll dp[5050];
ll mx[5050];
ll fact[MAXN] = {1, 1}, inv[MAXN] = {1, 1}, fact_inv[MAXN] = {1, 1};
void init() {
  for (ll i = 2; i < MAXN; i++) {
    fact[i] = ll(fact[i - 1]) * i % mod;
  }
  for (ll i = 2; i < MAXN; i++) {
    inv[i] = ll(inv[mod % i]) * (mod - mod / i) % mod;
  }
  for (ll i = 2; i < MAXN; i++) {
    fact_inv[i] = ll(fact_inv[i - 1]) * inv[i] % mod;
  }
}
ll nCr(ll n, ll r) {
  return n >= r && r >= 0
             ? ll(fact[n]) * fact_inv[n - r] % mod * fact_inv[r] % mod
             : 0;
}
ll nPr(ll n, ll r) {
  return n >= r && r >= 0 ? ll(fact[n]) * fact_inv[n - r] % mod : 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(20);
  init();
  ll n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 0) {
    cout << 1 << '\n';
    return 0;
  }
  memset(mx, -1, sizeof(mx));
  for (ll l = 0, r = -1, c = 0;;) {
    while (r < n - 1 && c + (s[r + 1] == '1') <= k) {
      ++r;
      c += s[r] == '1';
    }
    if (c < k) {
      break;
    }
    mx[l] = r;
    c -= s[l] == '1';
    ++l;
  }
  for (ll i = n - 1; i >= 0; i--) {
    if (mx[i] == -1) {
      dp[i] = 1;
    } else if (mx[i] == n - 1) {
      dp[i] = nCr(n - i, k);
    } else {
      dp[i] = dp[i + 1];
      ll R = mx[i];
      if (s[i] == '0') {
        dp[i] += nCr(R - i, k - 1);
        if (dp[i] >= mod) {
          dp[i] -= mod;
        }
      } else {
        dp[i] += nCr(R - i, k);
        if (dp[i] >= mod) {
          dp[i] -= mod;
        }
      }
    }
  }
  cout << dp[0] << '\n';
  return 0;
}

```