#include <bits/stdc++.h>
using namespace std;
const int N = 5010;
const int mod = 998244353;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10ll + ch - '0', ch = getchar();
  return x * f;
}
int n, k, pw[N], a[N];
char s[N];
void up(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int Pow(int x, int y) {
  if (!y) return 1;
  int t = Pow(x, y >> 1);
  t = (long long)t * t % mod;
  if (y & 1) t = (long long)t * x % mod;
  return t;
}
int fac[N], inv[N], fin[N];
void init() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = (long long)fac[i - 1] * i % mod;
  inv[0] = inv[1] = 1;
  for (int i = 2; i < N; i++)
    inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod;
  fin[0] = 1;
  for (int i = 1; i < N; i++) fin[i] = (long long)fin[i - 1] * inv[i] % mod;
}
int C(int n, int m) {
  if (n < m) return 0;
  return (long long)fac[n] * fin[m] % mod * fin[n - m] % mod;
}
int main() {
  n = read(), k = read();
  init();
  int l = 0;
  scanf("%s", s + 1);
  a[l++] = 0;
  for (int i = 1; i <= n; i++)
    if (s[i] == '1') a[l++] = i;
  a[l++] = n + 1;
  if (!k || k > l - 2) return puts("1"), 0;
  int ans = 0;
  for (int i = 0; i + 1 + k < l; i++) {
    up(ans, C(a[i + 1 + k] - a[i] - 1, k));
    if (i + 1 + k != l - 1)
      up(ans, mod - C(a[i + 1 + k] - a[i + 1] - 1, k - 1));
  }
  cout << ans;
}

### Prompt

Your challenge is to write a cpp solution 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;
const int N = 5010;
const int mod = 998244353;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10ll + ch - '0', ch = getchar();
  return x * f;
}
int n, k, pw[N], a[N];
char s[N];
void up(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int Pow(int x, int y) {
  if (!y) return 1;
  int t = Pow(x, y >> 1);
  t = (long long)t * t % mod;
  if (y & 1) t = (long long)t * x % mod;
  return t;
}
int fac[N], inv[N], fin[N];
void init() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = (long long)fac[i - 1] * i % mod;
  inv[0] = inv[1] = 1;
  for (int i = 2; i < N; i++)
    inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod;
  fin[0] = 1;
  for (int i = 1; i < N; i++) fin[i] = (long long)fin[i - 1] * inv[i] % mod;
}
int C(int n, int m) {
  if (n < m) return 0;
  return (long long)fac[n] * fin[m] % mod * fin[n - m] % mod;
}
int main() {
  n = read(), k = read();
  init();
  int l = 0;
  scanf("%s", s + 1);
  a[l++] = 0;
  for (int i = 1; i <= n; i++)
    if (s[i] == '1') a[l++] = i;
  a[l++] = n + 1;
  if (!k || k > l - 2) return puts("1"), 0;
  int ans = 0;
  for (int i = 0; i + 1 + k < l; i++) {
    up(ans, C(a[i + 1 + k] - a[i] - 1, k));
    if (i + 1 + k != l - 1)
      up(ans, mod - C(a[i + 1 + k] - a[i + 1] - 1, k - 1));
  }
  cout << ans;
}

```