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;

// G - G[i] хранит объект первых индексов каждого символа в строке s, начиная поиск с индекса i
// D - динамика, D[i][j] хранит минимальную длину x префикса строки s, в котором содержится a[:i] и b[:j]
// D[0][0] = 0 - то есть для пустых префиксов - минимальным будет пустой префикс s
const interleaving = (a, b, G) => {
    const D = Array.from({ length: a.length + 1 }, () => Array(b.length + 1).fill(Infinity)); D[0][0] = 0;
    for (let i = 0; i < a.length + 1; i++) {
        for (let j = 0; j < b.length + 1; j++) {
            if (!isFinite(D[i][j])) continue;
            const x = D[i][j];
            if (i !== a.length) D[i+1][j] = Math.min(D[i+1][j], G[x][ord(a[i])] + 1);
            if (j !== b.length) D[i][j+1] = Math.min(D[i][j+1], G[x][ord(b[j])] + 1);
        }
    }

    return isFinite(D[a.length][b.length]);
}

const problem = (s, p) => {
    const n = s.length;
    const G = Array.from({ length: n + 2}, () => Array(26).fill(Infinity));
    for (let i = n - 1; i > -1; i--) { G[i] = G[i + 1].slice(); G[i][ord(s[i])] = i; }
    for (let i = 0; i < p.length; i++) if (interleaving(p.slice(0, i), p.slice(i), G, n)) return 'YES';
    return 'NO';
}

let t = +readline()
while(t--) print(problem(readline(), readline()));
