#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 100;
const long long mod = 998244353;
long long t, n, m, k;
long long a[N];
string s;
vector<long long> pos;
void init() { return; }
long long quickpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
long long C(long long n, long long m) {
  if (n < m) return 0;
  long long a = 1, b = 1;
  m = min(m, n - m);
  for (long long i = 1; i <= m; i++) {
    (a *= n - i + 1) %= mod;
    (b *= i) %= mod;
  }
  return a % mod * quickpow(b, mod - 2) % mod;
}
void solve() {
  init();
  cin >> s;
  s = "." + s;
  pos.clear();
  pos.push_back(0);
  for (long long i = 1; i <= n; i++) {
    if (s[i] == '1') pos.push_back(i);
  }
  pos.push_back(n + 1);
  if (k == 0 || pos.size() - 2 < k) {
    cout << 1 << '\n';
    return;
  }
  long long ans = 0;
  for (long long l = k - 1; l <= k; l++) {
    long long sum = 0;
    for (long long i = k - l; i + k + 1 < pos.size(); i++) {
      sum = (sum + C(pos[i + l + 1] - 1 - pos[i], l)) % mod;
    }
    if ((k - l) % 2 == 0)
      ans = (ans + sum) % mod;
    else
      ans = (ans - sum + mod) % mod;
  }
  cout << ans << '\n';
  return;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  while (cin >> n >> k) {
    solve();
  }
  return 0;
}

### 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 long long N = 1e6 + 100;
const long long mod = 998244353;
long long t, n, m, k;
long long a[N];
string s;
vector<long long> pos;
void init() { return; }
long long quickpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
long long C(long long n, long long m) {
  if (n < m) return 0;
  long long a = 1, b = 1;
  m = min(m, n - m);
  for (long long i = 1; i <= m; i++) {
    (a *= n - i + 1) %= mod;
    (b *= i) %= mod;
  }
  return a % mod * quickpow(b, mod - 2) % mod;
}
void solve() {
  init();
  cin >> s;
  s = "." + s;
  pos.clear();
  pos.push_back(0);
  for (long long i = 1; i <= n; i++) {
    if (s[i] == '1') pos.push_back(i);
  }
  pos.push_back(n + 1);
  if (k == 0 || pos.size() - 2 < k) {
    cout << 1 << '\n';
    return;
  }
  long long ans = 0;
  for (long long l = k - 1; l <= k; l++) {
    long long sum = 0;
    for (long long i = k - l; i + k + 1 < pos.size(); i++) {
      sum = (sum + C(pos[i + l + 1] - 1 - pos[i], l)) % mod;
    }
    if ((k - l) % 2 == 0)
      ans = (ans + sum) % mod;
    else
      ans = (ans - sum + mod) % mod;
  }
  cout << ans << '\n';
  return;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  while (cin >> n >> k) {
    solve();
  }
  return 0;
}

```