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.util.*;
import java.io.*;
public class Solution {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    static void sort(int a[]){ // int -> long
        ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
        for(int i=0;i<a.length;i++)
        arr.add(a[i]);
        Collections.sort(arr);
        for(int i=0;i<a.length;i++)
        a[i]=arr.get(i);
        
    }
    private static long gcd(long a, long b){
        if(b==0)return a;
        return gcd(b,a%b);
    }
    private static long pow(long x,long y){
        if(y==0)return 1;
        long temp = pow(x, y/2);
        if(y%2==1){
            return (((x*temp)%mod)*temp)%mod;
        }
        else{
            return (temp*temp)%mod;
        }
    }
    static int log(long n){
        int res = 0;
        while(n>0){
            res++;
            n/=2;
        }
        return res;
    }

    static int mod = (int)998244353;
    static int INF = Integer.MAX_VALUE;
    static PrintWriter out;
    static FastReader sc ;
    public static void main(String[] args) throws IOException {
        sc = new FastReader();
        out = new PrintWriter(System.out);
        // primes();
        // ================================ //
        int n = sc.nextInt();
        int k = sc.nextInt();
        String s = sc.nextLine();
        solver(s, n, k);
        // ================================ //
        out.flush();
    }

    public static void solver(String s, int n, int k) {
        if(k==0){
            out.println( 1);
            return;
        }
        int[]pre = new int[n+1];
        for(int i=1;i<=n;i++){
            pre[i] = pre[i-1]+(s.charAt(i-1)-'0');
        }
        if(k>pre[n]){
            out.println( 1);
            return;
        }
        int start =0, end = 1;
        int last = 0;
        while(end<=n && pre[end]<=k){
            if(s.charAt(end-1)=='1')last = end;
            end++;
        }
        end--;
        long res = 0;
        // System.out.println("__"+ncr(10,2,mod));
        while(end<=n){
            if(start==0)
                res=(res+ncr(end-start, k, mod)+mod)%mod; 
            else
                res=(res+ncr(end-start, k, mod)-ncr(last-start-1,k-1,mod)+mod)%mod; 
            // System.out.println(start+" "+end+" "+last+"== "+res);
            // System.out.println("__"+    res);
            // if(res<0)res+=mod;
            
            int temp = end;
            while(end<=n && pre[end]-pre[start]<=k+1){
                if(end>0 && s.charAt(end-1)=='1')last = end;
                end++;
            }
            end--;
            if(temp==end)break;
            while(start<=end && pre[end]-pre[start]>k){
                start++;
            }
        }
        out.println( res);
    }

    // Returns n^(-1) mod p
    static long modInverse(long n, long p)
    {
        // System.out.println("__"+    pow(n,p-2));
        return pow(n, p - 2);
    }
 
    // Returns nCr % p using Fermat's
    // little theorem.
    static long ncr(int n, int r,
                             int p)
    {
 
          if (n<r)
              return 0;
      // Base case
        if (r == 0)
            return 1;
 
        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r
        long[] fac = new long[n + 1];
        fac[0] = 1;
 
        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;
 
        return ((long)fac[n] * modInverse(fac[r], p)
                % p * modInverse(fac[n - r], p)
                % p)
            % p;
    }
}
