#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10, M = 2 * N, mod = 998244353;
long long inv[N], fac[N], ifac[N];
void init(int n) {
  inv[1] = fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i <= n; i++) {
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
    fac[i] = fac[i - 1] * i % mod;
    ifac[i] = ifac[i - 1] * inv[i] % mod;
  }
}
long long C(int m, int n) { return fac[m] * ifac[n] % mod * ifac[m - n] % mod; }
int pre[N], fix[N];
string s;
int pos[N];
void solve() {
  int n, k;
  cin >> n >> k >> s;
  s = " " + s;
  int last = 1;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') pre[i] = last, last = i + 1;
  }
  last = n;
  for (int i = n; i >= 1; i--) {
    if (s[i] == '1') fix[i] = last, last = i - 1;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (s[i] == '1') pos[++cnt] = i;
  if (cnt < k || k == 0) {
    cout << 1 << '\n';
    return;
  }
  long long res = 0;
  for (int i = 1; i <= cnt - k + 1; i++) {
    int L = pre[pos[i]], R = fix[pos[i + k - 1]];
    int L1 = pre[pos[i + 1]], R1 = R;
    res += C(R - L + 1, k);
    if (i != cnt - k + 1) res -= C(R1 - L1 + 1, k - 1);
    res = (res + mod) % mod;
  }
  cout << res << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cout.precision(2), cout.setf(ios::fixed);
  init(1e4 + 10);
  solve();
  return 0;
}

### Prompt

Please formulate 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 = 3e5 + 10, M = 2 * N, mod = 998244353;
long long inv[N], fac[N], ifac[N];
void init(int n) {
  inv[1] = fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i <= n; i++) {
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
    fac[i] = fac[i - 1] * i % mod;
    ifac[i] = ifac[i - 1] * inv[i] % mod;
  }
}
long long C(int m, int n) { return fac[m] * ifac[n] % mod * ifac[m - n] % mod; }
int pre[N], fix[N];
string s;
int pos[N];
void solve() {
  int n, k;
  cin >> n >> k >> s;
  s = " " + s;
  int last = 1;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') pre[i] = last, last = i + 1;
  }
  last = n;
  for (int i = n; i >= 1; i--) {
    if (s[i] == '1') fix[i] = last, last = i - 1;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (s[i] == '1') pos[++cnt] = i;
  if (cnt < k || k == 0) {
    cout << 1 << '\n';
    return;
  }
  long long res = 0;
  for (int i = 1; i <= cnt - k + 1; i++) {
    int L = pre[pos[i]], R = fix[pos[i + k - 1]];
    int L1 = pre[pos[i + 1]], R1 = R;
    res += C(R - L + 1, k);
    if (i != cnt - k + 1) res -= C(R1 - L1 + 1, k - 1);
    res = (res + mod) % mod;
  }
  cout << res << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cout.precision(2), cout.setf(ios::fixed);
  init(1e4 + 10);
  solve();
  return 0;
}

```