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 inf = 1e9;
int test;
int n, m;
vector<vector<int>> nxt(402, vector<int>(26, inf));
int dp[402][402];
bool sol(string a, string b) {
  dp[0][0] = -1;
  int x = a.size();
  int y = b.size();
  for (int i = 0; i < (x + 1); i++)
    for (int j = 0; j < (y + 1); j++)
      if (i | j) {
        dp[i][j] = n;
        if (i) dp[i][j] = min(dp[i][j], nxt[dp[i - 1][j] + 1][a[i - 1] - 'a']);
        if (j) dp[i][j] = min(dp[i][j], nxt[dp[i][j - 1] + 1][b[j - 1] - 'a']);
      }
  return dp[x][y] < n;
}
void solve() {
  string s, t;
  cin >> s >> t;
  n = s.size();
  m = t.size();
  for (int i = n - 1; i >= 0; i--) {
    nxt[i] = nxt[i + 1];
    nxt[i][(int)s[i] - 'a'] = i;
  }
  for (int i = 0; i < (m); i++)
    if (sol(t.substr(0, i), t.substr(i))) {
      cout << "YES"
           << "\n";
      return;
    }
  cout << "NO"
       << "\n";
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> test;
  while (test--) solve();
  return 0;
}
