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 sys import stdin
  from itertools import accumulate
  input=stdin.readline
  rn=lambda:int(input())
  rns=lambda:map(int,input().split())
  rl=lambda:list(map(int,input().split()))
  rs=lambda:input().strip()
  YN=lambda x:print('YES') if x else print('NO')
  ceil_div=lambda a,b:-(-a//b)
  mod=998244353
  
  
  def make_nCr_mod(max_n=5000):
      max_n = min(max_n, mod - 1)
  
      fact, inv_fact = [0] * (max_n + 1), [0] * (max_n + 1)
      fact[0] = 1
      for i in range(max_n):
          fact[i + 1] = fact[i] * (i + 1) % mod
  
      inv_fact[-1] = pow(fact[-1], mod - 2, mod)
      for i in reversed(range(max_n)):
          inv_fact[i] = inv_fact[i + 1] * (i + 1) % mod
  
      def nCr_mod(n, r):
          res = 1
          while n or r:
              a, b = n % mod, r % mod
              if a < b:
                  return 0
              res = res * fact[a] % mod * inv_fact[b] % mod * inv_fact[a - b] % mod
              n //= mod
              r //= mod
          return res
  
      return nCr_mod
  
  
  
  n,k=rns()
  a=list(map(int,rs()))
  acc=list(accumulate(a))
  # print(acc)
  def solve():
      j=0
      ans=1
      choose=make_nCr_mod()
      first=True
      for i in range(n):
          # j+=1
          if j==n:
              break
          while j<n and acc[j]-acc[i]+a[i]<=k:
              if acc[j]-acc[i]+a[i]==k:
                  add=0
                  if not first:
                      if a[j]==0:
                          add=choose(j-i,k-1)
                      elif j-i+1>k:
                          add=choose(j-i,k)
                  else:
                      if a[j]==0:
                          add=choose(j-i+1,k-1)-1
                      elif j-i+1>k:
                          add=choose(j-i+1,k)-1
                      first=False
                  ans+=add
                  ans%=mod
                  # print(i,j,add)
              j+=1
      return ans
  print(solve())
  