#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
const int mod = 998244353;
void bmod(int &x) { x -= mod, x += x >> 31 & mod; }
int norm(int x) { return x < 0 ? x + mod : x; }
void Add(int &x, int y) { x += y - mod, x += x >> 31 & mod; }
void Sub(int &x, int y) { x -= y, x += x >> 31 & mod; }
int inv(int x, int k = mod - 2) {
  int r = 1;
  while (k) {
    if (k & 1) r = (long long)x * r % mod;
    x = (long long)x * x % mod;
    k >>= 1;
  }
  return r;
}
const int Flen = N;
int fact[Flen], infact[Flen];
void preF(int n = Flen - 1) {
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) fact[i] = (long long)fact[i - 1] * i % mod;
  infact[n] = inv(fact[n]);
  for (int i = n; i; --i) infact[i - 1] = (long long)infact[i] * i % mod;
}
int C(int x, int y) {
  return y < 0 || x < y
             ? 0
             : (long long)fact[x] * infact[y] % mod * infact[x - y] % mod;
}
int n, m;
char str[N];
int f[N], s[N];
int main() {
  preF();
  scanf("%d%d%s", &n, &m, str + 1);
  if (!m) return puts("1"), 0;
  f[0] = 1;
  for (int i = 1; i <= n; ++i) {
    s[i] = s[i - 1] + (str[i] -= 48);
  }
  for (int i = 1, j = 0; i <= n; ++i) {
    f[i] = f[i - 1];
    if (str[i]) {
      while (j < i && s[i] - s[j] > m) ++j;
      if (s[i] - s[j] != m) continue;
      if (s[j])
        Add(f[i], C(i - j - 1, m));
      else
        Add(f[i], C(i - j, m) - 1);
    } else {
      if (s[i] - s[j] != m) continue;
      Add(f[i], C(i - j - 1, m - 1));
    }
  }
  printf("%d\n", f[n]);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int N = 5005;
const int mod = 998244353;
void bmod(int &x) { x -= mod, x += x >> 31 & mod; }
int norm(int x) { return x < 0 ? x + mod : x; }
void Add(int &x, int y) { x += y - mod, x += x >> 31 & mod; }
void Sub(int &x, int y) { x -= y, x += x >> 31 & mod; }
int inv(int x, int k = mod - 2) {
  int r = 1;
  while (k) {
    if (k & 1) r = (long long)x * r % mod;
    x = (long long)x * x % mod;
    k >>= 1;
  }
  return r;
}
const int Flen = N;
int fact[Flen], infact[Flen];
void preF(int n = Flen - 1) {
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) fact[i] = (long long)fact[i - 1] * i % mod;
  infact[n] = inv(fact[n]);
  for (int i = n; i; --i) infact[i - 1] = (long long)infact[i] * i % mod;
}
int C(int x, int y) {
  return y < 0 || x < y
             ? 0
             : (long long)fact[x] * infact[y] % mod * infact[x - y] % mod;
}
int n, m;
char str[N];
int f[N], s[N];
int main() {
  preF();
  scanf("%d%d%s", &n, &m, str + 1);
  if (!m) return puts("1"), 0;
  f[0] = 1;
  for (int i = 1; i <= n; ++i) {
    s[i] = s[i - 1] + (str[i] -= 48);
  }
  for (int i = 1, j = 0; i <= n; ++i) {
    f[i] = f[i - 1];
    if (str[i]) {
      while (j < i && s[i] - s[j] > m) ++j;
      if (s[i] - s[j] != m) continue;
      if (s[j])
        Add(f[i], C(i - j - 1, m));
      else
        Add(f[i], C(i - j, m) - 1);
    } else {
      if (s[i] - s[j] != m) continue;
      Add(f[i], C(i - j - 1, m - 1));
    }
  }
  printf("%d\n", f[n]);
  return 0;
}

```