description: 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.
solution: import java.io.*;
  import java.util.*;
  
  public class CF1622D extends PrintWriter {
  	CF1622D() { super(System.out, true); }
  	Scanner sc = new Scanner(System.in);
  	public static void main(String[] $) {
  		CF1622D o = new CF1622D(); o.main(); o.flush();
  	}
  
  	static final int MD = 998244353;
  	void main() {
  		int n = sc.nextInt();
  		int k = sc.nextInt();
  		int[][] ch = new int[n + 1][k + 1];
  		ch[0][0] = 1;
  		for (int a = 1; a <= n; a++) {
  			ch[a][0] = 1;
  			for (int b = 1; b <= k; b++) 
  				ch[a][b] = (ch[a - 1][b] + ch[a - 1][b - 1]) % MD;
  		}
  		byte[] cc = sc.next().getBytes();
  		int[] qu = new int[n + 2];
  		int cnt = 0;
  		qu[cnt++] = 0;
  		for (int i = 1; i <= n; i++)
  			if (cc[i - 1] == '1')
  				qu[cnt++] = i;
  		qu[cnt++] = n + 1;
  		if (cnt - 2 < k || k == 0) {
  			println(1);
  			return;
  		}
  		long ans = 1;
  		for (int i = 1; i + k < cnt; i++) {
  			int a = qu[i + k] - qu[i - 1] - 1;
  			int b = qu[i + k] - qu[i] - 1;
  			ans += ch[a][k];
  			ans -= ch[b][k - 1];
  		}
  		for (int h = 1; h < k; h++) {
  			int a = qu[cnt - 1] - qu[cnt - 1 - h - 1] - 1;
  			int b = qu[cnt - 1] - qu[cnt - 1 - h] - 1;
  			ans += ch[a][h];
  			ans -= ch[b][h - 1];
  		}
  		ans %= MD;
  		if (ans < 0)
  			ans += MD;
  		println(ans);
  	}
  }
  