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 D {

	static long MOD = 998244353;
	static long[] factorial = new long[50001];
	
	public static void main(String[] args) throws IOException {

		factorial[0] = 1;
		
		for (long i = 1; i <= 5000; i++)
		    factorial[(int) i] = factorial[(int) (i - 1)] * i % MOD;
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		StringTokenizer st = new StringTokenizer(br.readLine());
		
		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		
		char[] ar = br.readLine().toCharArray();
		
		int l = -1;
		int r = -1;		
		int foc = 0;
		int roc = 0;
		int cnt = 0;
		boolean first = true;
		
		long tot = 0;
		
		for(int i = 0; i < n; i++) {
			
			if(ar[i] == '1') {
				
				if(cnt == k && l != -1) {
					
					foc = roc = 0;
					
					for(int j = r-1; j >= 0; j--) {
						
						if(ar[j] == '1' && j < l)
							break;
						
						if(ar[j] == '0')
							foc++;
						
					}
					
					for(int j = r+1; j < n; j++) {
						
						if(ar[j] == '1')
							break;
						
						roc++;
						
					}
					
					long add = bincoef(k+foc+roc, k) - ((first) ? 1 : bincoef(k+foc-1, k-1));
							
					if(add < 0)
						add += MOD;
							
					tot = (tot + add) % MOD;
					
					first = false;
															
					//System.out.println(l + " " + r + " " + foc + " " + roc + " " + tot);
											
				}
				
				if(l == -1) {
					
					l = i;
					r = i;
					
				}else
					r = i;
				
				cnt++;
				
				if(cnt > k) {
					
					for(int j = l+1; j <= r; j++) {
						
						if(ar[j] == '1') {
							
							l = j;
							break;
							
						}
						
					}
					
					cnt--;
					
				}
												
			}
			
		}
		
		if(cnt == k) {
			
			foc = roc = 0;
			
			for(int j = r-1; j >= 0; j--) {
				
				if(ar[j] == '1' && j < l)
					break;
				
				if(ar[j] == '0')
					foc++;
				
			}
			
			for(int j = r+1; j < n; j++) {
				
				if(ar[j] == '1')
					break;
				
				roc++;
				
			}
									
			long add = bincoef(k+foc+roc, k) - ((first) ? 1 : bincoef(k+foc-1, k-1));
			
			if(add < 0)
				add += MOD;
					
			tot = (tot + add) % MOD;
			
			//System.out.println(l + " " + r + " " + foc + " " + roc + " " + tot);
									
		}
		
		System.out.println((tot+1)%MOD);

	}
	
	static long bincoef(int n, int k) {
		
		if(n <= 0)
			return 1;
		
		if(k <= 0)
			return 1;
		
	    return factorial[n] * binpow(factorial[k] * factorial[n - k] % MOD, MOD - 2) % MOD;
	    
	}
	
	static long binpow(long a, long b) {
		
	    a %= MOD;
	    
	    long res = 1;
	    
	    while (b > 0) {
	    	
	        if ((b & 1) == 1)
	            res = res * a % MOD;
	        
	        a = a * a % MOD;
	        b >>= 1;
	        
	    }
	    
	    return res;
	    
	}

}
