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: /*
   * Everything is Hard 
   * Before Easy 
   * Jai Mata Dii 
   */ 
   
  import java.util.*;
  import java.io.*; 
    
  public class Main {
  	static class FastReader{ BufferedReader br;StringTokenizer st;public FastReader(){br = new BufferedReader(new InputStreamReader(System.in));}String next(){while (st == null || !st.hasMoreElements()){try{st = new StringTokenizer(br.readLine());}catch (IOException  e){e.printStackTrace();}}return st.nextToken();}int nextInt(){ return Integer.parseInt(next());}long nextLong(){return Long.parseLong(next());}double nextDouble(){return Double.parseDouble(next());}String nextLine(){String str = ""; try{str = br.readLine(); } catch (IOException e) {e.printStackTrace();} return str; }} 
  //	static long mod = (long)(1e9+7); 
  	 static long mod = 998244353; 
  //	 static Scanner sc = new Scanner(System.in); 
  	static FastReader sc = new FastReader(); 
  	static PrintWriter out = new PrintWriter(System.out);
  	static int ans;
  	public static void main (String[] args) {
  		int ttt = 1;
  //		ttt = sc.nextInt();
  		precompute();
  		z :for(int tc=1;tc<=ttt;tc++){
  			long nx = sc.nextLong();
  			long k = sc.nextLong();
  			char s[] = sc.next().toCharArray();
  			int i = 0, j = 0;
  			int cnt = 0, pre = 0;
  			long ans = 1;
  			boolean isPrev = false;
  			while(j<nx) {
  				if(s[j] == '1') {
  					cnt++;
  				}
  				if(cnt>k && k!=0) {
  //					System.out.println(i+" "+j);
  					long n = j - i;
  					long r = k;
  					
  					long cans = ((fac[(int)n] * power(fac[(int)(n-r)],mod-2,mod))%mod * power(fac[(int)r],mod-2,mod))%mod;
  					
  					ans = (ans + (cans-1L+mod)%mod)%mod;
  					
  					if(isPrev) {
  						n = pre - 1 - i + 1;
  						r = k - 1;
  						
  						cans = ((fac[(int)n] * power(fac[(int)(n-r)],mod-2,mod))%mod * power(fac[(int)r],mod-2,mod))%mod;
  						
  						ans = (ans - ((cans-1L+mod)%mod) + mod)%mod;
  					}
  					
  					isPrev = true;
  					pre = j;
  					
  					while(i<=j && cnt>k) {
  						if(s[i] == '1') cnt--;
  						i++;
  					}
  				}
  				j++;
  			}
  			if(cnt == k && i<=j && k>0) {
  //				System.out.println(i+" "+j);
  				long n = j - i;
  				long r = k;
  				
  				long cans = ((fac[(int)n] * power(fac[(int)(n-r)],mod-2,mod))%mod * power(fac[(int)r],mod-2,mod))%mod;
  				
  				ans = (ans + (cans-1L+mod)%mod)%mod;
  				
  				if(isPrev) {
  					n = pre - 1 - i + 1;
  					r = k - 1;
  					
  					cans = ((fac[(int)n] * power(fac[(int)(n-r)],mod-2,mod))%mod * power(fac[(int)r],mod-2,mod))%mod;
  					
  					ans = (ans - ((cans-1L+mod)%mod) + mod)%mod;
  				}
  			}
  			out.write(ans+"\n");
  		}
  		out.close();
  	}
  	
  	static long fac[] = new long[100006];
  //	static int mod = (int) (1e9+7);
  	
  	static void precompute() {
  		int n = 100005;
  		fac[0] = 1;
  		for(int i=1;i<=n;i++)
  			fac[i] = (i * fac[i-1])%mod;
  	}
  	
  	static int power(long base, long l, long mod2) {
  		
  		long ans = 1;
  		while(l!=0) {
  			if(l%2==0) {
  				base = (base*base)%mod2;
  				l/=2;
  			}
  			else {
  				ans = (ans*base)%mod2;
  				l-=1;
  			}
  		}
  		return (int) (ans%mod2);
  	}
  	
  	static long pow(long a, long b){long ret = 1;while(b>0){if(b%2 == 0){a = (a*a)%mod;b /= 2;}else{ret = (ret*a)%mod;b--;}}return ret%mod;}
  	static long gcd(long a,long b){if(b==0) return  a; return gcd(b,a%b); } 
  	private static void sort(int[] a) {List<Integer> k = new ArrayList<>();for(int val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  	private static void ini(List<Integer>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void init(List<int[]>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void sort(long[] a) {List<Long> k = new ArrayList<>();for(long val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  }