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;
using ll = long long;
using pii = pair<int, int>;
const int inf = 1e8 + 5;
int main() {
  int ct;
  string s, t;
  scanf("%d", &ct);
  while (ct--) {
    cin >> s >> t;
    int ns = int((s).size());
    int nt = int((t).size());
    vector<vector<int> > nextPosition(ns + 1, vector<int>(26, inf));
    for (int i = ns - 1; i >= 0; i--) {
      nextPosition[i] = nextPosition[i + 1];
      nextPosition[i][s[i] - 'a'] = i;
    }
    bool possible = false;
    for (int k = 0; k < nt; k++) {
      string left = t.substr(0, k);
      string right = t.substr(k);
      vector<vector<int> > dp(int((left).size()) + 1,
                              vector<int>(int((right).size()) + 1, inf));
      dp[0][0] = 0;
      for (int i = 0; i <= int((left).size()); i++) {
        for (int j = 0; j <= int((right).size()); j++) {
          if (dp[i][j] > int((s).size())) continue;
          int len = dp[i][j];
          if (i < int((left).size()) &&
              nextPosition[len][left[i] - 'a'] < inf) {
            dp[i + 1][j] =
                min(dp[i + 1][j], 1 + nextPosition[len][left[i] - 'a']);
          }
          if (j < int((right).size()) &&
              nextPosition[len][right[j] - 'a'] < inf) {
            dp[i][j + 1] =
                min(dp[i][j + 1], 1 + nextPosition[len][right[j] - 'a']);
          }
        }
      }
      if (dp[int((left).size())][int((right).size())] < inf) {
        possible = true;
        break;
      }
    }
    if (possible) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}
