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

import java.io.*;
import java.util.*;

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        String s[]=bu.readLine().split(" ");
        int n=Integer.parseInt(s[0]),k=Integer.parseInt(s[1]);

        char c[]=bu.readLine().toCharArray();
        ArrayList<Integer> one=new ArrayList<>();
        int i;
        for(i=0;i<n;i++)
        if(c[i]=='1') one.add(i);
        if(k==0 || one.size()<k) {System.out.println(1); return;}

        long f[]=new long[n+1],in[]=new long[n+1];
        f[0]=in[0]=1;
        for(i=1;i<=n;i++) f[i]=f[i-1]*i%M;
        in[n]=power(f[n],M-2);
        for(i=n-1;i>0;i--) in[i]=in[i+1]*(i+1)%M;

        long ans=0;
        for(i=0;i+k<=one.size();i++)
        {
            int z=0,o=k,j;
            for(j=i+1;j<i+k;j++)
            z+=one.get(j)-one.get(j-1)-1;
            if(i==0) z+=one.get(i);
            else z+=one.get(i)-one.get(i-1)-1;
            if(i+k<one.size()) z+=one.get(i+k)-one.get(i+k-1)-1;
            else z+=n-one.get(i+k-1)-1;

            //System.out.println(z+" "+o);
            long cur=f[z+o]*in[o]%M*in[z]%M;
            if(i>0)
            {
                o--;
                z=0;
                for(j=i;j<i+k;j++) z+=one.get(j)-one.get(j-1)-1;

                cur-=f[z+o]*in[o]%M*in[z]%M;
                cur=(cur+M)%M;
            }
            //System.out.println(cur);
            ans=(ans+cur)%M;
        }
        System.out.println(ans);
    }

    static long M=998244353;
    static long power(long a,long b)
    {
        long res=1;
        while(b!=0)
        {
            if(b%2==1) res=res*a%M;
            b>>=1;
            a=a*a%M;
        }
        return res;
    }
}