#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 9;
const long long MOD = 1e9 + 7;
const int mod = 998244353;
inline long long qpow(long long b, long long e, long long m) {
  long long a = 1;
  for (; e; e >>= 1, b = b * b % m)
    if (e & 1) a = a * b % m;
  return a;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y, y = z - y * (a / b);
  return d;
}
char s[maxn];
long long C[5009][5009];
void init() {
  C[0][0] = 1;
  for (int i = 1; i <= 5000; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      C[i][j] %= mod;
    }
  }
}
vector<int> ve;
int main() {
  init();
  int n, k;
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  if (k == 0) {
    cout << "1" << '\n';
    return 0;
  }
  ve.push_back(0);
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') ve.push_back(i);
  }
  ve.push_back(n + 1);
  if (ve.size() - 2 < k) {
    cout << 1 << '\n';
    return 0;
  }
  long long ans = 0;
  for (int i = 1;; i++) {
    if (i + k >= ve.size()) break;
    ans += C[ve[i + k] - ve[i - 1] - 1][k];
    if (i + k < (int)ve.size() - 1) ans -= C[ve[i + k] - ve[i] - 1][k - 1];
    ans %= mod;
  }
  ans = (ans % mod + mod) % mod;
  cout << ans << '\n';
}

### Prompt

Please create a solution in Cpp 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 maxn = 1e6 + 9;
const long long MOD = 1e9 + 7;
const int mod = 998244353;
inline long long qpow(long long b, long long e, long long m) {
  long long a = 1;
  for (; e; e >>= 1, b = b * b % m)
    if (e & 1) a = a * b % m;
  return a;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y, y = z - y * (a / b);
  return d;
}
char s[maxn];
long long C[5009][5009];
void init() {
  C[0][0] = 1;
  for (int i = 1; i <= 5000; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      C[i][j] %= mod;
    }
  }
}
vector<int> ve;
int main() {
  init();
  int n, k;
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  if (k == 0) {
    cout << "1" << '\n';
    return 0;
  }
  ve.push_back(0);
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') ve.push_back(i);
  }
  ve.push_back(n + 1);
  if (ve.size() - 2 < k) {
    cout << 1 << '\n';
    return 0;
  }
  long long ans = 0;
  for (int i = 1;; i++) {
    if (i + k >= ve.size()) break;
    ans += C[ve[i + k] - ve[i - 1] - 1][k];
    if (i + k < (int)ve.size() - 1) ans -= C[ve[i + k] - ve[i] - 1][k - 1];
    ans %= mod;
  }
  ans = (ans % mod + mod) % mod;
  cout << ans << '\n';
}

```