#include <bits/stdc++.h>
using namespace std;
long long Power(long long aaa, long long bbb) {
  long long result = 1;
  aaa %= 998244353;
  while (bbb) {
    if (bbb % 2 == 1) result *= aaa;
    aaa *= aaa;
    aaa %= 998244353;
    result %= 998244353;
    bbb /= 2;
  }
  return result;
}
long long Mod_Inv(long long aaa) {
  aaa %= 998244353;
  aaa = Power(aaa, 998244353 - 2);
  return aaa;
}
long long Factorial[1000001];
int Make_Factorial() {
  Factorial[0] = 1;
  for (long long i = 1; i < 1000001; i++) {
    Factorial[i] = i * Factorial[i - 1];
    Factorial[i] %= 998244353;
  }
  return 0;
}
int mk_fact = Make_Factorial();
vector<long long> fmoe(1000001);
int fmodf() {
  fmoe[1000001 - 1] = Mod_Inv(Factorial[1000001 - 1]);
  for (long long i = 1000001 - 2; i >= 0; i--) {
    fmoe[i] = fmoe[i + 1] * (i + 1);
    fmoe[i] %= 998244353;
  }
  return 0;
}
int fmoed = fmodf();
long long nCr(long long nnn, long long rrr) {
  if (nnn < rrr) return 0;
  long long aans = Factorial[nnn];
  aans %= 998244353;
  aans *= fmoe[rrr];
  aans %= 998244353;
  aans *= fmoe[nnn - rrr];
  aans %= 998244353;
  return aans;
}
long long nPr(long long nnn, long long rrr) {
  long long aans = Factorial[nnn];
  aans %= 998244353;
  aans *= fmoe[nnn - rrr];
  aans %= 998244353;
  return aans;
}
void sol() {
  long long n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 0) {
    cout << 1 << endl;
    return;
  }
  long long o = 0, z = 0;
  vector<long long> p;
  p.push_back(-1);
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') p.push_back(i);
  }
  p.push_back(n);
  int u = p.size();
  if (u - 2 < k) {
    cout << 1 << endl;
    return;
  }
  long long ans = 0;
  for (int i = 0; i + k < u - 1; i++) {
    long long n0 = (p[i + k + 1] - p[i] - 1) - k;
    long long tot = p[i + k + 1] - p[i] - 1;
    ans += nCr(tot, n0);
    ans %= 998244353;
  }
  for (int i = 1; i + k - 1 < u - 2; i++) {
    long long n0 = (p[i + k] - p[i] - 1) - (k - 1);
    long long tot = p[i + k] - p[i] - 1;
    ans += 998244353 - nCr(tot, n0);
    ans %= 998244353;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sol();
  return 0;
}

### 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;
long long Power(long long aaa, long long bbb) {
  long long result = 1;
  aaa %= 998244353;
  while (bbb) {
    if (bbb % 2 == 1) result *= aaa;
    aaa *= aaa;
    aaa %= 998244353;
    result %= 998244353;
    bbb /= 2;
  }
  return result;
}
long long Mod_Inv(long long aaa) {
  aaa %= 998244353;
  aaa = Power(aaa, 998244353 - 2);
  return aaa;
}
long long Factorial[1000001];
int Make_Factorial() {
  Factorial[0] = 1;
  for (long long i = 1; i < 1000001; i++) {
    Factorial[i] = i * Factorial[i - 1];
    Factorial[i] %= 998244353;
  }
  return 0;
}
int mk_fact = Make_Factorial();
vector<long long> fmoe(1000001);
int fmodf() {
  fmoe[1000001 - 1] = Mod_Inv(Factorial[1000001 - 1]);
  for (long long i = 1000001 - 2; i >= 0; i--) {
    fmoe[i] = fmoe[i + 1] * (i + 1);
    fmoe[i] %= 998244353;
  }
  return 0;
}
int fmoed = fmodf();
long long nCr(long long nnn, long long rrr) {
  if (nnn < rrr) return 0;
  long long aans = Factorial[nnn];
  aans %= 998244353;
  aans *= fmoe[rrr];
  aans %= 998244353;
  aans *= fmoe[nnn - rrr];
  aans %= 998244353;
  return aans;
}
long long nPr(long long nnn, long long rrr) {
  long long aans = Factorial[nnn];
  aans %= 998244353;
  aans *= fmoe[nnn - rrr];
  aans %= 998244353;
  return aans;
}
void sol() {
  long long n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 0) {
    cout << 1 << endl;
    return;
  }
  long long o = 0, z = 0;
  vector<long long> p;
  p.push_back(-1);
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') p.push_back(i);
  }
  p.push_back(n);
  int u = p.size();
  if (u - 2 < k) {
    cout << 1 << endl;
    return;
  }
  long long ans = 0;
  for (int i = 0; i + k < u - 1; i++) {
    long long n0 = (p[i + k + 1] - p[i] - 1) - k;
    long long tot = p[i + k + 1] - p[i] - 1;
    ans += nCr(tot, n0);
    ans %= 998244353;
  }
  for (int i = 1; i + k - 1 < u - 2; i++) {
    long long n0 = (p[i + k] - p[i] - 1) - (k - 1);
    long long tot = p[i + k] - p[i] - 1;
    ans += 998244353 - nCr(tot, n0);
    ans %= 998244353;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sol();
  return 0;
}

```