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 sys
input = sys.stdin.readline
test = int(input())
for _ in range(test):
  s = input().rstrip()
  t = input().rstrip()
  n = len(s)
  m = len(t)
  ansls = []
  pos = [[1000 for i in range(26)] for j in range(n+2)]
  for i in range(n+1)[::-1]:
    if i < n:
      for j in range(26):
        pos[i][j] = pos[i+1][j]
    if i > 0:
      x = ord(s[i-1])-97
      pos[i][x] = i
  flg = 0
  for i in range(m):
    t1 = t[:i]
    t2 = t[i:]
    m1 = len(t1)
    m2 = len(t2)
    dp = [[1000 for i in range(m2+1)] for j in range(m1+1)]
    dp[0][0] = 0
    for j in range(m1+1):
      for k in range(m2+1):
        if j > 0 and dp[j-1][k] < 1000:
          t1x = ord(t1[j-1])-97
          dp[j][k] = min(dp[j][k],pos[dp[j-1][k]+1][t1x])
        if k > 0 and dp[j][k-1] < 1000:
          t2x = ord(t2[k-1])-97
          dp[j][k] = min(dp[j][k],pos[dp[j][k-1]+1][t2x])
    if dp[-1][-1] < 1000:
      flg = 1
      break
  if flg:
    print("YES")
  else:
    print("NO")