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 _1622_D {
      static long MOD = 998244353;
      static long[] fact = new long[5005];
      static long[] inv_fact = new long[5005];
      public static void main(String[] args) throws IOException {
          long cur1 = 1;
          long cur2 = 1;
          for(int i = 0; i < 5005; i++) {
              fact[i] = cur1;
              inv_fact[i] = cur2;
              cur1 = modmult(cur1, i + 1);
              cur2 = modmult(cur2, modinv(i + 1));
          }
          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter out = new PrintWriter(System.out);
          StringTokenizer line = new StringTokenizer(in.readLine());
          int n = Integer.parseInt(line.nextToken());
          int k = Integer.parseInt(line.nextToken());
          String s = in.readLine();
          int total = 0;
          for(int i = 0; i < n; i++) {
              if(s.charAt(i) == '1') {
                  total++;
              }
          }
          long res = 1;
          if(k > 0) res = modadd(res, calc(n, k, s, k, total));
          if(k > 1) res = modadd(res, -calc(n, k, s, k - 1, total));
          out.println(res);
          in.close();
          out.close();
      }  
      static long calc(int n, int k, String s, int w, int total) {
          long res = 0;
          int r = 0;
          int count = 0;
          int ones = 0;
          for(int i = 0; i < n; i++) {
              if(i == 0 || s.charAt(i - 1) == '1') {
                  if(i > 0) {
                      count--;
                      ones++;
                  }
                  while(r < n) {
                      if(s.charAt(r) == '1') {
                          count++;
                          if(count > w) {
                              count--;
                              break;
                          }
                      }
                      r++;
                  }
                  if(ones >= k - w && count == w && total - (ones + w) >= k - w) {
                      res = modadd(res, binom(r - i, w) - 1);
                  }
              }
          }
          return res;
      }
      static long binom(int n, int k) {
          return modmult(modmult(fact[n], inv_fact[n - k]), inv_fact[k]);
      }
      static long modadd(long a, long b) {
          return (a + b + MOD) % MOD;
      }
      static long modmult(long a, long b) {
          return a * b % MOD;
      }
      static long modinv(long a) {
          return binpow(a, MOD - 2);
      }
      static long binpow(long a, long b) {
          if(b == 0) {
              return 1;
          }
          long small = binpow(a, b / 2);
          if(b % 2 == 0) {
              return modmult(small, small);
          }else {
              return modmult(modmult(small, small), a);
          }
      }
  }
  