#include <bits/stdc++.h>
using namespace std;
const int N = 2 * 1e5 + 10;
const int mod = 998244353;
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int mul(int a, int b) { return a * 1ll * b % mod; }
long long f[N];
long long fp(long long b, long long p) {
  if (p == 0) return 1;
  return fp(b * b % mod, p / 2) * (p & 1 ? b : 1) % mod;
}
long long inv(long long x) { return fp(x, mod - 2); }
long long nCk(long long n, long long k) {
  return f[n] * inv(f[n - k]) % mod * inv(f[k]) % mod;
}
void Ineedhelp() {
  f[0] = 1;
  for (int i = 1; i < N; ++i) {
    f[i] = f[i - 1] * i % mod;
  }
  int n, k;
  string s;
  cin >> n >> k >> s;
  int i = 0, j = -1, cnt1 = 0, last1 = -1;
  while (j < n - 1 && cnt1 + (s[j + 1] == '1') <= k) {
    ++j;
    cnt1 += s[j] == '1';
    if (s[j] == '1') last1 = j;
  }
  if (cnt1 < k || k == 0) {
    cout << "1\n";
    return;
  }
  long long ways = nCk(j - i + 1, k);
  while (j < n - 1) {
    int jj = j;
    ++j;
    while (j < n - 1 && s[j + 1] == '0') ++j;
    while (i < j && s[i] == '0') ++i;
    ++i;
    ways = ways - nCk(jj - i + 1, k - 1) + mod;
    ways %= mod;
    ways += nCk(j - i + 1, k);
    ways %= mod;
  }
  cout << ways << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) Ineedhelp();
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 2 * 1e5 + 10;
const int mod = 998244353;
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int mul(int a, int b) { return a * 1ll * b % mod; }
long long f[N];
long long fp(long long b, long long p) {
  if (p == 0) return 1;
  return fp(b * b % mod, p / 2) * (p & 1 ? b : 1) % mod;
}
long long inv(long long x) { return fp(x, mod - 2); }
long long nCk(long long n, long long k) {
  return f[n] * inv(f[n - k]) % mod * inv(f[k]) % mod;
}
void Ineedhelp() {
  f[0] = 1;
  for (int i = 1; i < N; ++i) {
    f[i] = f[i - 1] * i % mod;
  }
  int n, k;
  string s;
  cin >> n >> k >> s;
  int i = 0, j = -1, cnt1 = 0, last1 = -1;
  while (j < n - 1 && cnt1 + (s[j + 1] == '1') <= k) {
    ++j;
    cnt1 += s[j] == '1';
    if (s[j] == '1') last1 = j;
  }
  if (cnt1 < k || k == 0) {
    cout << "1\n";
    return;
  }
  long long ways = nCk(j - i + 1, k);
  while (j < n - 1) {
    int jj = j;
    ++j;
    while (j < n - 1 && s[j + 1] == '0') ++j;
    while (i < j && s[i] == '0') ++i;
    ++i;
    ways = ways - nCk(jj - i + 1, k - 1) + mod;
    ways %= mod;
    ways += nCk(j - i + 1, k);
    ways %= mod;
  }
  cout << ways << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) Ineedhelp();
}

```