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 + 7;
const int MOD = 1e9 + 7;
const int dx[4] = {0, 0, -1, 1};
const int dy[4] = {-1, 1, 0, 0};
const int N = 402;
int n, m;
string s, t;
vector<int> pos[N];
int dp[N][N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    n = s.size();
    m = t.size();
    s = '#' + s;
    t = '#' + t;
    for (int i = 0; i < 26; i++) pos[i + 'a'].clear();
    for (int i = 1; i <= n; i++) pos[s[i]].push_back(i);
    bool found = false;
    for (int k = 1; k <= m; k++) {
      if (found) break;
      string a = "#";
      string b = "#";
      for (int i = 1; i <= k; i++) a += t[i];
      for (int i = k + 1; i <= m; i++) b += t[i];
      int x = a.size() - 1;
      int y = b.size() - 1;
      for (int i = 0; i <= x; i++) {
        for (int j = 0; j <= y; j++) {
          dp[i][j] = INF;
        }
      }
      dp[0][0] = 0;
      for (int i = 0; i <= x; i++) {
        for (int j = 0; j <= y; j++) {
          if (dp[i][j] != INF) {
            if (i + 1 <= x) {
              int p = upper_bound(pos[a[i + 1]].begin(), pos[a[i + 1]].end(),
                                  dp[i][j]) -
                      pos[a[i + 1]].begin();
              if (p != pos[a[i + 1]].size())
                dp[i + 1][j] = min(dp[i + 1][j], pos[a[i + 1]][p]);
            }
            if (j + 1 <= y) {
              int p = upper_bound(pos[b[j + 1]].begin(), pos[b[j + 1]].end(),
                                  dp[i][j]) -
                      pos[b[j + 1]].begin();
              if (p != pos[b[j + 1]].size())
                dp[i][j + 1] = min(dp[i][j + 1], pos[b[j + 1]][p]);
            }
          }
        }
      }
      if (dp[x][y] <= n) found = true;
    }
    cout << (found ? "YES" : "NO") << '\n';
  }
  return 0;
}
