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: from functools import lru_cache
  
  @lru_cache(None)
  def fac(x):
      if x == 0:
          return 1
      else:
          return x * fac(x - 1) % MOD
          
  @lru_cache(None)
  def ifac(x):
      return pow(fac(x), MOD - 2, MOD)
  
  def C(x, y):
      if x < y:
          return 0
      return fac(x) * ifac(y) * ifac(x - y) % MOD
  
  
  MOD = 998244353
  for i in range(5000):
      fac(i)
  n, k = map(int, input().split())
  s = input()
  if k == 0:
      print(1)
  else:
      left = 0
      cnt = 0
      res = 0
      start = False
      flag = False
      for i in range(len(s)): 
          if s[i] == '1':
              cnt += 1
          while cnt > k:
              if s[left] == '1':
                  cnt -= 1
              left += 1
          if cnt == k:
              if not start:
                  res += C(i - left + 1, k)
              else:
                  if s[i] == '1':
                      res += C(i - left, k)
                  else:
                      res += C(i - left, k - 1)
              start = True
              flag = True
              res %= MOD
      if res == 0 and not flag:
          res = 1
      print(res)
  
              