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;
int INF = int(1e9);
string s, t;
bool can(string &a, string &b, vector<vector<int>> &nxt) {
  vector<vector<int>> dp(int(a.size()) + 1,
                         vector<int>(int(b.size()) + 1, INF));
  dp[0][0] = 0;
  for (int i = 0, thx = int(a.size()) + 1; i < thx; i++)
    for (int j = 0, thx = int(b.size()) + 1; j < thx; j++) {
      if (dp[i][j] > int(s.size())) continue;
      int len = dp[i][j];
      if (i + 1 <= int(a.size()) && nxt[len][a[i]] < INF) {
        dp[i + 1][j] = min(dp[i + 1][j], nxt[len][a[i]] + 1);
      }
      if (j + 1 <= int(b.size()) && nxt[len][b[j]] < INF) {
        dp[i][j + 1] = min(dp[i][j + 1], nxt[len][b[j]] + 1);
      }
    }
  return dp[int(a.size())][int(b.size())] < INF;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    for (auto &c : s) {
      c -= 'a';
    }
    for (auto &c : t) {
      c -= 'a';
    }
    vector<vector<int>> nxt(int(s.size()) + 1, vector<int>(26, INF));
    for (int i = int(s.size()) - 1; i >= 0; i--) {
      nxt[i] = nxt[i + 1];
      nxt[i][s[i]] = i;
    }
    bool cann = false;
    for (int i = 0, thx = int(t.size()); i < thx; i++) {
      string a = t.substr(0, i), b = t.substr(i, int(t.size()));
      if (can(a, b, nxt)) {
        cann = true;
        break;
      }
    }
    if (cann)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
