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 math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
  fact=[1]*5005
  for i in range(1,5005):
  	fact[i]=(fact[i-1]*i)%mod2
  n,k=M();s=S()
  if k==0 or k>s.count('1'):
  	print(1)
  else:
  	ans=0;a=[]
  	for i in range(n):
  		if s[i]=='1':a.append(i)
  	for i in range(len(a)-k+1):
  		l=a[i]-1;r=a[i+k-1]+1;p=r-1
  		while l>=0 and s[l]=='0':l-=1
  		while r<n and s[r]=='0':r+=1
  		l+=1;r-=1;t=(r-l+1)
  		f1=(fact[t]*pow(fact[t-k],mod2-2,mod2)*pow(fact[k],mod2-2,mod2))%mod2
  		ans=(ans+f1)%mod2
  		t1=t-(r-p+1);k1=k-1
  		if i==0:continue
  		f2=(fact[t1]*pow(fact[t1-k1],mod2-2,mod2)*pow(fact[k1],mod2-2,mod2))%mod2
  		ans=(ans-f2)%mod2
  	print(ans)