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
'use strict'
const ord = x => x.charCodeAt(0) - 97;

const problem = (s, p) => {
    const n = s.length;
    const F = [];
    for (let i = 0; i < n + 2; i++) F[i] = Array(26).fill(n);
    for (let i = n-1; i > -1; i--) {
        F[i] = F[i + 1].slice()
        F[i][ord(s[i])] = i
    }
    const interleaving = (l, r) => {
        const dp = Array(r.length + 1).fill(n); dp[0] = -1;

        for (let j = 1; j < r.length + 1; j++) {
            dp[j] = F[dp[j - 1] + 1][ord(r[j - 1])]
        }

        for (let i = 1; i < l.length + 1; i++) {
            dp[0] = F[dp[0] + 1][ord(l[i - 1])]

            for (let j = 1; j < r.length + 1; j++) {
                const a = F[dp[j] + 1][ord(l[i - 1])]
                const b = F[dp[j - 1] + 1][ord(r[j - 1])]
                dp[j] = Math.min(a, b)
            }
        }
        return dp[dp.length - 1] < n
    }

    for (let i = 0; i < p.length; i++) if (interleaving(p.slice(0, i), p.slice(i))) return 'YES'
    return 'NO'
}

let t = +readline()
while(t--) print(problem(readline().split(''), readline()));
