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
#include <bits/stdc++.h>
using namespace std;
const int N = 440;
const int Inf = 1e9;
string s, t;
int Dp[2][N][N];
void Solve() {
  cin >> s >> t;
  int n = s.size(), m = t.size();
  for (int j1 = 0; j1 <= m; j1++)
    for (int j2 = 0; j2 <= m; j2++) Dp[n & 1][j1][j2] = (j2 == m ? j1 : -Inf);
  for (int i = n - 1; ~i; --i) {
    for (int j1 = 0; j1 <= m; j1++) {
      for (int j2 = 0; j2 <= m; j2++) {
        Dp[i & 1][j1][j2] = Dp[(i + 1) & 1][j1][j2];
        if (j1 < t.size() && t[j1] == s[i])
          Dp[i & 1][j1][j2] =
              max(Dp[i & 1][j1][j2], Dp[(i + 1) & 1][j1 + 1][j2]);
        if (j2 < t.size() && t[j2] == s[i])
          Dp[i & 1][j1][j2] =
              max(Dp[i & 1][j1][j2], Dp[(i + 1) & 1][j1][j2 + 1]);
      }
    }
  }
  bool Ok = false;
  for (int j2 = 0; j2 <= t.size(); j2++)
    if (Dp[0 & 1][0][j2] >= j2) Ok = true;
  puts(Ok ? "YES" : "NO");
}
int main() {
  int T;
  cin >> T;
  while (T--) Solve();
}
