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.
5 2
00011
 5 4
11011
 81 5
100000101000000011001101010100001010111000000110000000001000101101111000001000110
 4 3
1011
 47 25
11110011111001110100101010011000110111011011101
 291 159
011001110111110110001111000110001001111100010001011000110100110010001101100110111010101111100111110000110111001100111010001000011001010011011100101000101011010110010011010100100101111010111000001011110101001011101110100100111110011001101010001110110101010101100011101110111000011111011001110
 459 15
101010101010101010101010101010101010101010101111111111111111111111010101010101010101010101011111111111111111111101010101010101010101111111111111111111111101010101010101010111111111111111111111111101010101010101111111111111111111111111110110101010101011111111111111111111111111111001000111111111111111111111111111110000011111111111111111111111111111000001111111111111111111111111111100001111111111111111111111111111100111111111111111111111111111110111111111111
 105 58
110101000111100000011011011101110100001000110011010101100010111001011111010010111101011011010110011011110
 52 11
0110011111001000010000100010010001010100000100110100

10
 5
 187538
 4
 314445058
 1
 0
 132732930
 478323746

#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][5050];
ll cnt[5050][5050];
ll mn[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;
}
void add(ll &x, ll y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
  if (x < 0) {
    x += mod;
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(20);
  init();
  ll n, k;
  cin >> n >> k;
  if (k == 0) {
    cout << 1 << '\n';
    return 0;
  }
  string s;
  cin >> s;
  for (ll i = 0; i < n; i++) {
    cnt[i][i] = s[i] == '1';
  }
  for (ll i = 0; i < n; i++) {
    for (ll j = i + 1; j < n; j++) {
      cnt[i][j] = cnt[i][j - 1] + cnt[j][j];
    }
  }
  memset(mn, -1, sizeof(mn));
  for (ll i = 0; i < n; i++) {
    for (ll j = i; j >= 0; j--) {
      if (cnt[j][i] == k) {
        mn[i] = j;
      }
    }
  }
  for (ll l = 1; l <= n; l++) {
    for (ll i = 0; i + l <= n; i++) {
      ll j = i + l - 1;
      if (cnt[i][j] < k) {
        dp[i][j] = 1;
      } else if (cnt[i][j] == k) {
        dp[i][j] = nCr(l, k);
      } else if (cnt[i][j] == j - i + 1) {
        dp[i][j] = 1;
      } else {
        dp[i][j] = dp[i][j - 1];
        char r = s[j];
        ll L = mn[j];
        if (r == '0') {
          add(dp[i][j],
              (i <= L - 1 ? dp[i][L - 1] : 1) * nCr(j - L, k - 1) % mod);
        } else {
          add(dp[i][j], (i <= L - 1 ? dp[i][L - 1] : 1) * nCr(j - L, k) % mod);
        }
      }
      0xF;
    }
  }
  cout << dp[0][n - 1] << '\n';
  return 0;
}
