You are given two strings s and t. The string s consists of lowercase Latin letters and at most one wildcard character '*', the string t consists only of lowercase Latin letters. The length of the string s equals n, the length of the string t equals m.

The wildcard character '*' in the string s (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of s can be replaced with anything. If it is possible to replace a wildcard character '*' in s to obtain a string t, then the string t matches the pattern s.

For example, if s="aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".

If the given string t matches the given string s, print "YES", otherwise print "NO".

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of the string s and the length of the string t, respectively.

The second line contains string s of length n, which consists of lowercase Latin letters and at most one wildcard character '*'.

The third line contains string t of length m, which consists only of lowercase Latin letters.

Output

Print "YES" (without quotes), if you can obtain the string t from the string s. Otherwise print "NO" (without quotes).

Examples

Input

6 10
code*s
codeforces


Output

YES


Input

6 5
vk*cup
vkcup


Output

YES


Input

1 1
v
k


Output

NO


Input

9 6
gfgf*gfgf
gfgfgf


Output

NO

Note

In the first example a wildcard character '*' can be replaced with a string "force". So the string s after this replacement is "codeforces" and the answer is "YES".

In the second example a wildcard character '*' can be replaced with an empty string. So the string s after this replacement is "vkcup" and the answer is "YES".

There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".

In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string t so the answer is "NO".
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
import java.util.StringTokenizer;

public class A {
	public static void main(String[] args) throws IOException {
		FastScanner in= new FastScanner(System.in);
		PrintWriter out= new PrintWriter(System.out);
		
		int n= in.nextInt();
		int m= in.nextInt();
		
		String s= in.next();
		String t= in.next();
		
		if(n>m+1) {
			System.out.println("NO");
			return;
		}
		
		int ind= s.indexOf("*");
		if(ind==-1) {
			if(s.equals(t)) {
				System.out.println("YES");
				return;
			}
			System.out.println("NO");
			return;
		}
		
		if(t.startsWith(s.substring(0, ind)) && t.endsWith(s.substring(ind + 1))) {
			System.out.println("YES");
			return;
		}
		System.out.println("NO");
		
	}
	static class p implements Comparable<p>{
		int x;
		int y;
		public p(int a, int b) {
			x=a;
			y=b;
		}
		public int compareTo(p o) {
			if(o.x-this.x==0) return this.y-o.y;
			return o.x-this.x;
		}
	}
	static class FenwickTree {
		long[] ft;
		public FenwickTree(int n) {
			ft = new long[n + 1];
		}

		int rsq(int b) {
			int sum = 0;
			for (; b > 0; b -= (b & (-b)))
				sum += ft[b];
			return sum;
		}

		int rsq(int a, int b) {
			return rsq(b) - (a == 1 ? 0 : rsq(a - 1));
		}

		void update(int k, int v) {
			for (; k < ft.length; k += (k & (-k)))
				ft[k] += v;
		}
	}
	public static long gcd(long a, long b) {
		return b==0 ? a: gcd(b, a%b);
	}
	public static long lcm(long a, long b) {
		return a*b/ gcd(a,b);
	}
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
			st = new StringTokenizer("");
		}

		public String next() throws IOException {
			if (!st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
				return next();
			}
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
		public double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	}
}
/*
6
ATATA
ATATAT
CCCCC
ACTGGGACT
ACTG
CCCAGGG
UIBPMUIBQKAQABPMLWWZIVLSMGBWBPMAKQMVKMA
ATHINGOFBEAUTYISAJOYFOREVER
CUFTDDYZXTLAFDKVSFLTVUFFXF
*/