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.
5 2
00011
 5 4
11011
 81 5
100000101000000011001101010100001010111000000110000000001000101101111000001000110
 4 3
1011
 47 25
11110011111001110100101010011000110111011011101
 291 159
011001110111110110001111000110001001111100010001011000110100110010001101100110111010101111100111110000110111001100111010001000011001010011011100101000101011010110010011010100100101111010111000001011110101001011101110100100111110011001101010001110110101010101100011101110111000011111011001110
 459 15
101010101010101010101010101010101010101010101111111111111111111111010101010101010101010101011111111111111111111101010101010101010101111111111111111111111101010101010101010111111111111111111111111101010101010101111111111111111111111111110110101010101011111111111111111111111111111001000111111111111111111111111111110000011111111111111111111111111111000001111111111111111111111111111100001111111111111111111111111111100111111111111111111111111111110111111111111
 105 58
110101000111100000011011011101110100001000110011010101100010111001011111010010111101011011010110011011110
 52 11
0110011111001000010000100010010001010100000100110100

10
 5
 187538
 4
 314445058
 1
 0
 132732930
 478323746

# cook your dish here
import sys
import bisect
from bisect import bisect_left as lb
from bisect import bisect_right as rb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
sys.setrecursionlimit(3*10**5)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs2(root,par):
    if par!=-1:
        dp[root]=dp[par]+1
    for i in range(1,20):
        if kthpar[root][i-1]!=-1:
            kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
    for child in gp[root]:
        if child==par:continue
        kthpar[child][0]=root
        dfs(child,root)
        
ans=0
a=[]
n,k=0,0
b=[]
vis=[]
tot=0
time=[]
time1=[]
adj=[]
mx=-1
eps=0.0000001
fact=[0]*(6010)
inv=[0]*(6010)
fact[0]=1
fact[1]=1
inv[0]=1
inv[1]=1
for i in range(2,6002):
    fact[i]=(fact[i-1]*i)%mod1
    inv[i]=pow(fact[i],mod1-2,mod1)
def ncr(n,r):
    if r==0 or n==0:
        return 1
    ans=fact[n]*inv[r]*inv[n-r]
    ans%=mod1
    return ans
def hnbhai(tc):
    ##O(n) done 
    n,k=sd()
    s="s"+sa()
    prefix=[0]*(n+1)
    suffix=[0]*(n+1)
    dp=[[0,0] for i in range(n+1)]
    tot=0
    pos=[]
    for i in range(1,n+1):
        dp[i]=dp[i-1].copy()
        dp[i][int(s[i])]+=1
        if int(s[i])==0: tot+=1
        else:
            pos.append(i)
            prefix[i]=tot
            tot=0
    if k==0 or dp[n][1]-dp[0][1]<k:
        print(1)
        return
    temp=0
    for i in range(n,0,-1):
        if int(s[i])==0:temp+=1
        else:
            suffix[i]=temp
            temp=0
    tot=0
    ans=0
    l,r=100000,-1
    for i in range(len(pos)):
        if i+k==len(pos):
            temp=prefix[pos[i]]+n-pos[i]+1
            ans+=ncr(temp,k)
            ans%=mod1
            break
        temp=prefix[pos[i]]+pos[i+k]-pos[i]
        ans+=ncr(temp,k)
        ans-=ncr(temp-1-prefix[pos[i]],k-1)
        ans%=mod1
    print(ans)
    return
for _ in range(1):
    hnbhai(_+1)
