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
from sys import stdin, stdout


def solve(s1, s2, next):
    # i => s1, j => s2
    dp = [[INF for _ in range(len(s2)+1)] for _ in range(len(s1)+1)]
    dp[0][0] = 0
    for i in range(len(s1)+1):
        for j in range(len(s2)+1):
            if dp[i][j] == INF:
                continue

            if i < len(s1) and dp[i][j] < len(next) and next[dp[i][j]][ord(s1[i]) - ord('a')] < INF:
                dp[i+1][j] = min(dp[i+1][j], next[dp[i][j]][ord(s1[i]) - ord('a')] + 1)
            if j < len(s2) and dp[i][j] < len(next) and next[dp[i][j]][ord(s2[j]) - ord('a')] < INF:
                dp[i][j+1] = min(dp[i][j+1], next[dp[i][j]][ord(s2[j]) - ord('a')] + 1)

    return dp[len(s1)][len(s2)]


INF = 1e20
T = int(stdin.readline())
for _ in range(T):
    s = stdin.readline().strip()
    rs = stdin.readline().strip()
    next = [[INF for _ in range(26)] for _ in range(len(s))]

    for i in range(len(s)-1, -1, -1):
        if i < len(s)-1:
            for j in range(26):
                next[i][j] = next[i+1][j]
        next[i][ord(s[i]) - ord('a')] = i

    found = False

    if len(rs) == 1:
        if rs in s:
            found = True
    else:
        for p in range(1, len(rs)):
            s1 = rs[:p]
            s2 = rs[p:]

            if solve(s1, s2, next) < INF:
                found = True
                break

    if found:
        stdout.write('YES\n')
    else:
        stdout.write('NO\n')

