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.util.*;
  import java.io.*;
  import java.lang.reflect.Array;
  
  public class Main {
  	public static final int N = 5005, mod = 998244353;
  	public static int n, k;
  	public static char[] s = new char[N];
  	public static int Pow(long a, long b) {
  		long ans = 1;
  		while(b != 0) {
  			if((b & 1) != 0) ans = ans * a % mod;
  			a = a * a % mod;
  			b >>= 1;
  		}
  		return (int) (ans % mod);
  	}
  	public static int[] fac = new int[N], ifac = new int[N];
  	public static void init(int n) {
  		fac[0] = 1;
  		for(int i = 1; i<=n; i++) fac[i] = (int) (((long)fac[i-1] * i) % mod);
  		ifac[n] = Pow(fac[n], mod-2);
  		for(int i=n-1; i>=0; i--) ifac[i] = (int) ((long) ifac[i+1]*(i+1)%mod);
  	}
  	public static int C(int a, int b) {
  		if(a<b||b<0) return 0;
  		return (int) (((long)fac[a])*ifac[b]%mod*ifac[a-b]%mod);
  	}
  	public static void main(String[] args) throws IOException {
  		//scanf("%d%d%s", &n, &k, s+1);
  		BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer(input.readLine());
  		n = Integer.parseInt(st.nextToken());
  		k = Integer.parseInt(st.nextToken());
  		
  		String tttemp = input.readLine();
  		for(int i = 1;i<=n;i++) {
  			s[i] = tttemp.charAt(i - 1);
  		}
  		
  		if(k == 0) {
  			System.out.println(1);
  			return;
  		}
  		init(n);
  		int cnt = 0, ans = 0;
  		for(int i=1, j=1; i<=n; i++) {
  			if(s[i]=='0') continue;
  			++cnt;
  			if(cnt<k) continue;
  			boolean f = false;
  			while(cnt>k) {
  				f = true;
  				cnt -= s[j] - '0';
  				++j;
  			}
  			if(f) ans = (ans - C(i-j, k-1) + mod)%mod;
  			++i;
  			while(i<=n && s[i]=='0') ++i;
  			--i;
  			ans = (ans + C(i-j+1, k))%mod;
  		}
  		if(cnt<k) ans = 1;
  		System.out.println(ans);
  }
  }
  