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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  int inf = 524;
  while (t--) {
    string s, t;
    cin >> s >> t;
    vector<vector<int>> nex(inf + 3, vector<int>(150, inf));
    for (int a = s.size() - 1; a >= 0; a--) {
      for (int b = 0; b < 150; b++) {
        nex[a][b] = nex[a + 1][b];
      }
      nex[a][s[a]] = a + 1;
    }
    bool good = 0;
    int n = t.size();
    string l = t, r = "";
    for (int a = 0; a < n; a++) {
      int ls = l.size();
      int rs = r.size();
      vector<vector<int>> dp(ls + 3, vector<int>(rs + 3, inf));
      dp[0][0] = 0;
      for (int i = 0; i <= ls; i++) {
        for (int j = 0; j <= rs; j++) {
          if (!i && !j) continue;
          if (i == 0) {
            dp[i][j] = nex[dp[i][j - 1]][r[j - 1]];
          } else if (j == 0) {
            dp[i][j] = nex[dp[i - 1][j]][l[i - 1]];
          } else {
            dp[i][j] =
                min(nex[dp[i][j - 1]][r[j - 1]], nex[dp[i - 1][j]][l[i - 1]]);
          }
        }
      }
      if (dp[ls][rs] != inf) {
        good = 1;
        break;
      }
      r = l.back() + r;
      l.pop_back();
    }
    if (good) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}
