#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long n, k;
long long fac[5010] = {1, 1};
long long inv[5010] = {1, 1};
char s[5010];
vector<long long> v;
long long qpow(long long a, long long b, long long res = 1) {
  for (; b; b >>= 1, (a *= a) %= mod)
    if (b & 1) (res *= a) %= mod;
  return res;
}
void init(void) {
  for (long long i = 2; i <= 5000; i++) fac[i] = fac[i - 1] * i % mod;
  for (long long i = 2; i <= 5000; i++) inv[i] = qpow(fac[i], mod - 2);
  return;
}
long long C(long long a, long long b) {
  return fac[a] * inv[b] % mod * inv[a - b] % mod;
}
void solve(long long res = 0) {
  scanf("%lld %lld %s", &n, &k, s + 1);
  if (!k) {
    puts("1");
    return;
  }
  s[0] = s[n + 1] = '1';
  v.clear();
  for (long long i = 0; i <= n + 1; i++)
    if (s[i] == '1') v.push_back(i);
  if (v.size() - 2 < k) {
    puts("1");
    return;
  }
  res = C(v[k + 1] - 1, k);
  for (long long i = k + 1; i < (long long)v.size() - 1; i++) {
    long long len = v[i + 1] - v[i - k] - 1;
    long long cov = v[i] - v[i - k] - 1;
    (res += C(len, k)) %= mod;
    (res -= C(cov, k - 1)) %= mod;
    (res += mod) %= mod;
  }
  printf("%lld\n", res);
  return;
}
signed main() {
  init();
  solve();
  return 0;
}

### Prompt

In cpp, your task is to solve 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 long long mod = 998244353;
long long n, k;
long long fac[5010] = {1, 1};
long long inv[5010] = {1, 1};
char s[5010];
vector<long long> v;
long long qpow(long long a, long long b, long long res = 1) {
  for (; b; b >>= 1, (a *= a) %= mod)
    if (b & 1) (res *= a) %= mod;
  return res;
}
void init(void) {
  for (long long i = 2; i <= 5000; i++) fac[i] = fac[i - 1] * i % mod;
  for (long long i = 2; i <= 5000; i++) inv[i] = qpow(fac[i], mod - 2);
  return;
}
long long C(long long a, long long b) {
  return fac[a] * inv[b] % mod * inv[a - b] % mod;
}
void solve(long long res = 0) {
  scanf("%lld %lld %s", &n, &k, s + 1);
  if (!k) {
    puts("1");
    return;
  }
  s[0] = s[n + 1] = '1';
  v.clear();
  for (long long i = 0; i <= n + 1; i++)
    if (s[i] == '1') v.push_back(i);
  if (v.size() - 2 < k) {
    puts("1");
    return;
  }
  res = C(v[k + 1] - 1, k);
  for (long long i = k + 1; i < (long long)v.size() - 1; i++) {
    long long len = v[i + 1] - v[i - k] - 1;
    long long cov = v[i] - v[i - k] - 1;
    (res += C(len, k)) %= mod;
    (res -= C(cov, k - 1)) %= mod;
    (res += mod) %= mod;
  }
  printf("%lld\n", res);
  return;
}
signed main() {
  init();
  solve();
  return 0;
}

```