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;
const long long MAX = 2000000;
const long long MOD = 998244353;
const long long OO = 0x3f3f3f3f;
const double EPS = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long cnt;
long long dp[5005][5005];
long long seen[5005][5005];
long long C(long long n, long long k) {
  if (k > n) return 0;
  if (seen[n][k]) return dp[n][k];
  if (n == k or k == 0) return 1;
  long long r = (C(n - 1, k - 1) + C(n - 1, k)) % MOD;
  seen[n][k] = 1;
  return dp[n][k] = r;
}
void solve() {
  long long n, k;
  string s;
  cin >> n >> k >> s;
  vector<long long> ones;
  ones.push_back(0);
  for (long long i = 1; i <= n; ++i)
    if (s[i - 1] == '1') ones.push_back(i);
  ones.push_back(n + 1);
  if ((long long)ones.size() - 2 < k or k == 0) {
    cout << "1\n";
    return;
  }
  vector<pair<long long, long long> > nck;
  for (long long i = 1; i + k < (long long)ones.size(); ++i) {
    long long l = ones[i - 1] + 1;
    long long r = ones[i + k] - 1;
    nck.push_back({l, r});
  }
  long long ans = 0;
  for (long long i = 0; i < (long long)nck.size(); ++i) {
    long long l = nck[i].first, r = nck[i].second;
    ans = (ans + C(r - l + 1, k)) % MOD;
    if (i) {
      long long m = nck[i - 1].second - l + 1;
      if (m > 0) ans = (ans - C(m, k - 1) + MOD) % MOD;
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  for (long long caso = 1; caso <= t; ++caso) {
    solve();
  }
  return 0;
}
