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 Main {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter pr = new PrintWriter(System.out);
	static String readLine() throws IOException {
		return br.readLine();
	}
	static String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(readLine());
		return st.nextToken();
	}
	static int readInt() throws IOException {
		return Integer.parseInt(next());
	}
	static long readLong() throws IOException {
		return Long.parseLong(next());
	}
	static double readDouble() throws IOException {
		return Double.parseDouble(next());
	}
	static char readChar() throws IOException {
		return next().charAt(0);
	}
	static class Pair implements Comparable<Pair> {
		int f, s;
		Pair(int f, int s) {
			this.f = f;
			this.s = s;
		}
		public int compareTo(Pair other) {
			if (this.f != other.f) return this.f - other.f;
			return this.s - other.s;
		}
	}
	final static int mod = 998244353;
	static long pow(long b, long p) {
		long ret = 1;
		while (p > 0) {
			if ((p&1) == 1) ret = ret*b%mod;
			b = b*b%mod;
			p >>= 1;
		}
		return ret;
	}
	static long fac[], inv[];
	static long C(int n, int k) {
		return fac[n]*(inv[k]*inv[n - k]%mod)%mod;
	}
	static void solve() throws IOException {
		int n = readInt(), k = readInt();
		fac = new long[n + 1];
		inv = new long[n + 1];
		fac[0] = inv[0] = 1;
		for (int i = 1; i <= n; ++i) {
			fac[i] = fac[i - 1]*i%mod;
			inv[i] = pow(fac[i], mod - 2);
		}
		String s = " " + readLine();
		int tot = 0;
		for (int i = 1; i <= n; ++i)
			tot += s.charAt(i) - '0';
		if (k == 0 || tot < k) {
			pr.println(1);
			return;
		}
		long ans = 1;
		for (int L = 1; L <= n; ++L) {
			int R = L - 1, cnt = 0;
			while (R < n && cnt + s.charAt(R + 1) - '0' <= k) {
				cnt += s.charAt(R + 1) - '0';
				++R;
			}
			int num = R - L + 1;
			if (cnt == 0 || cnt == num) continue;
			long add = 0;
			if (s.charAt(L) == '1') add = C(num - 1, cnt);
			else add = C(num - 1, cnt - 1);
			ans = (ans + add)%mod;
		}
		pr.println(ans);
	}
	public static void main(String[] args) throws IOException {
		solve();
		//for (int t = readInt(); t > 0; --t) solve();
		pr.close();
	}
}
