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 sys
  import io, os
  input = sys.stdin.readline
  
  mod = 998244353
  N = 5001
  
  fac = [1]*(N+1)
  finv = [1]*(N+1)
  for i in range(N):
      fac[i+1] = fac[i] * (i+1) % mod
  finv[-1] = pow(fac[-1], mod-2, mod)
  for i in reversed(range(N)):
      finv[i] = finv[i+1] * (i+1) % mod
  
  def cmb1(n, r):
      #if r <0 or r > n:
          #return 0
      #r = min(r, n-r)
      return (fac[n] * finv[r] * finv[n-r]) % mod
  
  n, k = map(int, input().split())
  s = str(input().rstrip())
  
  C = [0]*(n+1)
  for i, c in enumerate(s):
      if c == '1':
          C[i+1] = C[i]+1
      else:
          C[i+1] = C[i]
  if C[n] < k:
      print(1)
      exit()
  ans = 1
  for i in range(n):
      if s[i] == '0':
          flag = True
      else:
          flag = False
      for j in range(i+1, n):
          cnt1 = C[j+1]-C[i]
          cnt = j-i+1
          if cnt1 > k:
              break
          cnt -= 2
          if flag:
              cnt1 -= 1
          if s[j] == '0':
              cnt1 -= 1
          if cnt1 <= cnt and cnt1 >= 0 and cnt >= 0:
              ans += cmb1(cnt, cnt1)
              ans %= mod
  print(ans)
  