You have a string s and a chip, which you can place onto any character of this string. 

After placing the chip, you move it to the right several (maybe zero) times, i. e. you perform the following operation several times: if the current position of the chip is i, you move it to the position i + 1. Of course, moving the chip to the right is impossible if it is already in the last position.

After moving the chip to the right, you move it to the left several (maybe zero) times, i. e. you perform the following operation several times: if the current position of the chip is i, you move it to the position i - 1. Of course, moving the chip to the left is impossible if it is already in the first position.

When you place a chip or move it, you write down the character where the chip ends up after your action. For example, if s is abcdef, you place the chip onto the 3-rd character, move it to the right 2 times and then move it to the left 3 times, you write down the string cdedcb.

You are given two strings s and t. Your task is to determine whether it's possible to perform the described operations with s so that you write down the string t as a result.

Input

The first line contains one integer q (1 ≤ q ≤ 500) — the number of test cases.

Each test case consists of two lines. The first line contains the string s (1 ≤ |s| ≤ 500), the second line contains the string t (1 ≤ |t| ≤ 2 ⋅ |s| - 1). Both strings consist of lowercase English characters.

It is guaranteed that the sum of |s| over all test cases does not exceed 500.

Output

For each test case, print "YES" if you can obtain the string t by performing the process mentioned in the statement with the string s, or "NO" if you cannot.

You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).

Example

Input


6
abcdef
cdedcb
aaa
aaaaa
aab
baaa
ab
b
abcdef
abcdef
ba
baa


Output


YES
YES
NO
YES
YES
NO

Note

Consider the examples.

The first test case is described in the statement.

In the second test case, you can place the chip on the 1-st position, move it twice to the right, and then move it twice to the left.

In the fourth test case, you can place the chip on the 2-nd position, and then don't move it at all.

In the fifth test case, you can place the chip on the 1-st position, move it 5 times to the right, and then finish the process.
import java.io.*;
import java.util.*;


public class Main2{
	static Boolean[][][]memo;
	static char[]s,t;
	static boolean dp(int right,int i,int j) {
		if(j==t.length)return true;
		if(i<0 || i>=s.length)return false;
		if(memo[right][i][j]!=null)return memo[right][i][j];
		if(s[i]!=t[j])return memo[right][i][j]=false;
		if(right==1) {
			return memo[right][i][j]=dp(right, i+1, j+1) || dp(0, i-1, j+1);
		}
		return memo[right][i][j]=dp(0, i-1, j+1);
	}
	static void main() throws Exception{
		s=sc.nextLine().toCharArray();
		t=sc.nextLine().toCharArray();
		memo=new Boolean[2][s.length][t.length];
		boolean can=false;
		for(int i=0;i<s.length;i++) {
			can|=dp(1, i, 0);
		}
		pw.println(can?"YES":"NO");
	}
	public static void main(String[] args) throws Exception{
		sc=new MScanner(System.in);
		pw = new PrintWriter(System.out);
		int tc=1;
        tc=sc.nextInt();
		for(int i=1;i<=tc;i++) {
//			pw.printf("Case #%d: ", i);
			main();
		}
		
		pw.flush();
	}
	static PrintWriter pw;
	static MScanner sc;
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
	 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
	 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
	 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
	 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
	 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
	 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	 
		public boolean ready() throws IOException {
			return br.ready();
		}
	 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
		
	}
	static void dbg(int[]in) {
		System.out.println(Arrays.toString(in));
	}
	static void dbg(long[]in) {
		System.out.println(Arrays.toString(in));
	}
	static void sort(int[]in) {
		shuffle(in);
		Arrays.sort(in);
	}
	static void sort(long[]in) {
		shuffle(in);
		Arrays.sort(in);
	}
	static void shuffle(int[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			int tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
	static void shuffle(long[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			long tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
}