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.*;
  
  public class D {
  
      static FastScanner fs = new FastScanner();
      static Scanner scn = new Scanner(System.in);
      static PrintWriter pw = new PrintWriter(System.out);
      static StringBuilder sb = new StringBuilder("");
      final static long MOD = 998244353;
      final static int INF = 1000000000;
  
      public static void main(String[] args) {
          int n = fs.nextInt(), k = fs.nextInt();
          String s = fs.next();
          int[][] binom = new int[n+1][n+1];
          for (int i = 0; i <= n; i++) {
              binom[i][0] = 1;
              for (int j = 1; j < i; j++) {
                  binom[i][j] = binom[i-1][j-1] + binom[i-1][j];
                  if (binom[i][j] >= MOD) binom[i][j] -= MOD;
              }
              binom[i][i] = 1;
          }
          int cnt = 0;
          for (int i = 0; i < n; i++) if (s.charAt(i) == '1') cnt++;
          if (cnt < k) sb.append("1\n");
          else {
              long ans = 0;
              for (int i = 0; i < n; i++) {
                  int cnt0 = 0, cnt1 = 0;
                  for (int j = i; j < n; j++) {
                      if (cnt1 == k && s.charAt(j) == '1') break;
                      if (s.charAt(j) == '0') cnt0++;
                      if (s.charAt(j) == '1') cnt1++;
                  }
                  if (s.charAt(i) == '0' && cnt1 >= 1) ans += binom[cnt0 + cnt1 - 1][cnt0];
                  if (s.charAt(i) == '1' && cnt0 >= 1) ans += binom[cnt0 + cnt1 - 1][cnt1];
              }
              ans++;
              ans %= MOD;
              sb.append(ans + "\n");
          }
          pw.print(sb.toString());
          pw.close();
      }
      
      static void sort(int[] a) {
          ArrayList<Integer> al = new ArrayList<>();
          for (int i : a)
              al.add(i);
          Collections.sort(al);
          for (int i = 0; i < a.length; i++)
              a[i] = al.get(i);
      }
  
      static class FastScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
          String next() {while (!st.hasMoreTokens()) try {st = new StringTokenizer(br.readLine());} catch (IOException e) {}return st.nextToken();}
          int nextInt() {return Integer.parseInt(next());}
          long nextLong() {return Long.parseLong(next());}
          double nextDouble() {return Double.parseDouble(next());}
      }
  }