#include <bits/stdc++.h>
template <typename Sep, typename Head, typename... Tail>
void fold_print(Sep sep, Head&& h, Tail&&... args) {
  ((std::cerr << h) << ... << (std::cerr << sep, args));
}
std::mt19937_64 myrand(
    std::chrono::high_resolution_clock::now().time_since_epoch().count());
using namespace std;
using LL = long long int;
void init();
void process();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = 1;
  init();
  for (int t = 1; t <= T; t++) {
    process();
  }
  cout.flush();
  return 0;
}
constexpr LL MOD = 998244353;
vector<LL> fac(5000 * 2 + 1), inv(5000 * 2 + 1);
LL fastpw(LL x, LL p) {
  if (p == 0) return 1;
  LL tmp = fastpw(x * x % MOD, p / 2);
  if (p % 2 == 0)
    return tmp;
  else
    return tmp * x % MOD;
}
LL modinv(LL x) { return fastpw(x, MOD - 2); }
void init() {
  fac[0] = 1;
  for (LL i = 1; i < fac.size(); i++) {
    fac[i] = i * fac[i - 1] % MOD;
  }
  for (size_t i = 0; i < inv.size(); i++) {
    inv[i] = modinv(fac[i]);
  }
}
LL C(LL n, LL m) { return fac[n] * inv[m] % MOD * inv[n - m] % MOD; }
void process() {
  int n, k;
  cin >> n >> k;
  vector<int> arr(n);
  {
    for (auto& x : arr) {
      char ch;
      cin >> ch;
      x = ch - '0';
    }
  }
  auto permute = [&](LL l, LL r, LL one) {
    LL len = r - l + 1;
    LL cnt[] = {len - one, one};
    if (len == 1 || cnt[0] == len || cnt[1] == len) return 0ll;
    cnt[1 - arr[l]] -= 1;
    return C(len - 1, cnt[0]);
  };
  LL ans = 1, r = 0, cnt = arr[0];
  for (LL i = 0; i < n; cnt -= arr[i], i += 1) {
    while (r + 1 < n && cnt + arr[r + 1] <= k) {
      cnt += arr[r + 1], r += 1;
    }
    if (cnt != k) break;
    if (r == n - 1) {
      for (int j = i; j < n; cnt -= arr[j], j += 1) {
        ans = (ans + permute(j, r, cnt)) % MOD;
      }
      break;
    } else {
      ans = (ans + permute(i, r, cnt)) % MOD;
    }
  }
  cout << ans << '\n';
}

### 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>
template <typename Sep, typename Head, typename... Tail>
void fold_print(Sep sep, Head&& h, Tail&&... args) {
  ((std::cerr << h) << ... << (std::cerr << sep, args));
}
std::mt19937_64 myrand(
    std::chrono::high_resolution_clock::now().time_since_epoch().count());
using namespace std;
using LL = long long int;
void init();
void process();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = 1;
  init();
  for (int t = 1; t <= T; t++) {
    process();
  }
  cout.flush();
  return 0;
}
constexpr LL MOD = 998244353;
vector<LL> fac(5000 * 2 + 1), inv(5000 * 2 + 1);
LL fastpw(LL x, LL p) {
  if (p == 0) return 1;
  LL tmp = fastpw(x * x % MOD, p / 2);
  if (p % 2 == 0)
    return tmp;
  else
    return tmp * x % MOD;
}
LL modinv(LL x) { return fastpw(x, MOD - 2); }
void init() {
  fac[0] = 1;
  for (LL i = 1; i < fac.size(); i++) {
    fac[i] = i * fac[i - 1] % MOD;
  }
  for (size_t i = 0; i < inv.size(); i++) {
    inv[i] = modinv(fac[i]);
  }
}
LL C(LL n, LL m) { return fac[n] * inv[m] % MOD * inv[n - m] % MOD; }
void process() {
  int n, k;
  cin >> n >> k;
  vector<int> arr(n);
  {
    for (auto& x : arr) {
      char ch;
      cin >> ch;
      x = ch - '0';
    }
  }
  auto permute = [&](LL l, LL r, LL one) {
    LL len = r - l + 1;
    LL cnt[] = {len - one, one};
    if (len == 1 || cnt[0] == len || cnt[1] == len) return 0ll;
    cnt[1 - arr[l]] -= 1;
    return C(len - 1, cnt[0]);
  };
  LL ans = 1, r = 0, cnt = arr[0];
  for (LL i = 0; i < n; cnt -= arr[i], i += 1) {
    while (r + 1 < n && cnt + arr[r + 1] <= k) {
      cnt += arr[r + 1], r += 1;
    }
    if (cnt != k) break;
    if (r == n - 1) {
      for (int j = i; j < n; cnt -= arr[j], j += 1) {
        ans = (ans + permute(j, r, cnt)) % MOD;
      }
      break;
    } else {
      ans = (ans + permute(i, r, cnt)) % MOD;
    }
  }
  cout << ans << '\n';
}

```