#include <bits/stdc++.h>
using namespace std;
template <typename T>
void rd(T& x) {
  int f = 0, c;
  while (!isdigit(c = getchar())) f ^= !(c ^ 45);
  x = (c & 15);
  while (isdigit(c = getchar())) x = x * 10 + (c & 15);
  if (f) x = -x;
}
template <typename T>
void pt(T x, int c = 10) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) pt(x / 10, -1);
  putchar(x % 10 + 48);
  if (c != -1) putchar(c);
}
template <typename T>
void umax(T& x, const T& y) {
  if (x < y) x = y;
}
template <typename T>
void umin(T& x, const T& y) {
  if (x > y) x = y;
}
using vi = vector<int>;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<long, long>;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 998244353;
int n, k;
string s;
ll jc[maxn], inv[maxn];
int sum[maxn];
ll qp(ll a, ll b) {
  ll ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ans;
}
ll calc(ll x, ll y) { return jc[x] * inv[y] % mod * inv[x - y] % mod; }
ll calc2(int l, int r) { return calc(r - l + 1, sum[r] - sum[l - 1]); }
int main() {
  cin >> n >> k >> s;
  if (k == 0) {
    cout << 1 << "\n";
    return 0;
  }
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % mod;
  inv[n] = qp(jc[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; i++) sum[i] = (s[i - 1] == '1') + sum[i - 1];
  ll ans = 1;
  int pr = 1;
  for (int l = 1, r = 1; r <= n;) {
    while (r < n && sum[r + 1] - sum[l - 1] <= k) r++;
    if (sum[r] - sum[l - 1] < k) break;
    ans += calc2(l, r) - calc2(l, pr);
    pr = r;
    while (sum[r] - sum[l - 1] == k) l++;
  }
  cout << (ans % mod + mod) % mod << "\n";
  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;
template <typename T>
void rd(T& x) {
  int f = 0, c;
  while (!isdigit(c = getchar())) f ^= !(c ^ 45);
  x = (c & 15);
  while (isdigit(c = getchar())) x = x * 10 + (c & 15);
  if (f) x = -x;
}
template <typename T>
void pt(T x, int c = 10) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) pt(x / 10, -1);
  putchar(x % 10 + 48);
  if (c != -1) putchar(c);
}
template <typename T>
void umax(T& x, const T& y) {
  if (x < y) x = y;
}
template <typename T>
void umin(T& x, const T& y) {
  if (x > y) x = y;
}
using vi = vector<int>;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<long, long>;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 998244353;
int n, k;
string s;
ll jc[maxn], inv[maxn];
int sum[maxn];
ll qp(ll a, ll b) {
  ll ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ans;
}
ll calc(ll x, ll y) { return jc[x] * inv[y] % mod * inv[x - y] % mod; }
ll calc2(int l, int r) { return calc(r - l + 1, sum[r] - sum[l - 1]); }
int main() {
  cin >> n >> k >> s;
  if (k == 0) {
    cout << 1 << "\n";
    return 0;
  }
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % mod;
  inv[n] = qp(jc[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; i++) sum[i] = (s[i - 1] == '1') + sum[i - 1];
  ll ans = 1;
  int pr = 1;
  for (int l = 1, r = 1; r <= n;) {
    while (r < n && sum[r + 1] - sum[l - 1] <= k) r++;
    if (sum[r] - sum[l - 1] < k) break;
    ans += calc2(l, r) - calc2(l, pr);
    pr = r;
    while (sum[r] - sum[l - 1] == k) l++;
  }
  cout << (ans % mod + mod) % mod << "\n";
  return 0;
}

```