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;
vector<pair<int, int> > kandydaci[130];
bool czy[405][405];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s1, s2;
    cin >> s1 >> s2;
    int sz2 = s2.size();
    bool yes = 0;
    for (int p = 0; p < (int)s2.size(); p++) {
      string a = "", b = "";
      for (int i = 0; i < p; i++) a += s2[i];
      for (int i = p; i < (int)s2.size(); i++) b += s2[i];
      if ((int)a.size() > 0) kandydaci[(int)a[0]].push_back({1, 0});
      if ((int)b.size() > 0) kandydaci[(int)b[0]].push_back({0, 1});
      for (int i = 1; i <= (int)s1.size() && yes == 0; i++) {
        int c = s1[i - 1];
        vector<pair<int, int> > toAdd[130];
        for (int j = 0; j < (int)kandydaci[c].size(); j++) {
          pair<int, int> x = kandydaci[c][j];
          if (x.first + x.second == sz2) {
            yes = 1;
            break;
          }
          if (x.first + 1 <= (int)a.size() && czy[x.first + 1][x.second] == 0) {
            toAdd[(int)a[x.first]].push_back({x.first + 1, x.second});
            czy[x.first + 1][x.second] = 1;
          }
          if (x.second + 1 <= (int)b.size() &&
              czy[x.first][x.second + 1] == 0) {
            toAdd[(int)b[x.second]].push_back({x.first, x.second + 1});
            czy[x.first][x.second + 1] = 1;
          }
        }
        kandydaci[c].clear();
        for (int j = 97; j <= 122; j++) {
          for (int k = 0; k < (int)toAdd[j].size(); k++) {
            pair<int, int> x = toAdd[j][k];
            kandydaci[j].push_back(x);
          }
        }
      }
      for (int i = 97; i <= 122; i++) kandydaci[i].clear();
      if (yes) break;
      for (int i = 0; i <= 400; i++)
        for (int j = 0; j <= 400; j++) czy[i][j] = 0;
    }
    if (yes)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
