You are given a string s. You can build new string p from s using the following operation no more than two times: 

  1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|; 
  2. erase the chosen subsequence from s (s can become empty); 
  3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}). 



Of course, initially the string p is empty. 

For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.

Can you build a given string t using the algorithm above?

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.

The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.

It's guaranteed that the total length of strings s doesn't exceed 400.

Output

Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.

Example

Input


4
ababcd
abcba
a
b
defi
fed
xyz
x


Output


YES
NO
NO
YES
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskF solver = new TaskF();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskF {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int T = in.nextInt();
            iter:
            while (T-- > 0) {
                char[] s = in.next().toCharArray();
                char[] t = in.next().toCharArray();
                for (int i = 0; i <= t.length; i++) {
                    if (existsDisjoint(s, i, t)) {
                        out.println("YES");
                        continue iter;
                    }
                }
                out.println("NO");
            }
        }

        boolean existsDisjoint(char[] s, int split, char[] t) {
            char[] s1 = new char[split];
            char[] s2 = new char[t.length - split];
            for (int i = 0; i < split; i++) {
                s1[i] = t[i];
            }
            for (int i = split; i < t.length; i++) {
                s2[i - split] = t[i];
            }
            int[][] dp = new int[s.length + 1][s1.length + 1];
            for (int[] dp1 : dp) {
                Arrays.fill(dp1, Integer.MIN_VALUE);
            }
            dp[0][0] = 0;
            for (int i = 1; i <= s.length; i++) {
                for (int j = 0; j <= s1.length; j++) {
                    dp[i][j] = dp[i - 1][j];
                    if (j > 0 && s[i - 1] == s1[j - 1]) {
                        dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 1]);
                    }
                    if (dp[i - 1][j] < s2.length && dp[i - 1][j] >= 0) {
                        dp[i][j] = Math.max(dp[i][j], dp[i - 1][j] + (s2[dp[i - 1][j]] == s[i - 1] ? 1 : 0));
                    }

                }
            }
            return (dp[s.length][s1.length] == s2.length);
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

