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
# -*- coding: utf-8 -*-

import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
from typing import List

"""
created by shhuan at 2020/3/19 13:35

"""


def check(s, a, b, after):
    ns, na, nb = len(s), len(a), len(b)
    if ns < na + nb:
        return False

    dp = [[0 for _ in range(nb+1)] for _ in range(na+1)]
    for i in range(na+1):
        for j in range(nb+1):
            if i == 0 and j == 0:
                continue
            dp[i][j] = min(after[dp[i-1][j]][a[i-1]] if i > 0 else ns, after[dp[i][j-1]][b[j-1]] if j > 0 else ns) + 1

    return dp[na][nb] <= ns


def solve(s, t):
    ns = len(s)
    after = [[ns for _ in range(26)] for _ in range(ns+2)]
    for i in range(ns-1, -1, -1):
        for j in range(26):
            after[i][j] = after[i+1][j]
        after[i][s[i]] = i

    for i in range(len(t)):
        a, b = t[:i], t[i:]
        if check(s, a, b, after):
            return 'YES'

    return 'NO'


T = int(input())
ans = []
for i in range(T):
    s = input()
    t = input()
    s = [ord(v) - ord('a') for v in s]
    t = [ord(v) - ord('a') for v in t]
    ans.append(solve(s, t))

print('\n'.join(ans))