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;
template <typename T>
void print(T a) {
  for (auto x : a) cout << x << ' ';
  cout << endl;
}
const int INF = 1e9 + 9;
void solve() {
  string s, t;
  cin >> s >> t;
  int n = (int)(s).size();
  vector<vector<int> > nxt(n + 1, vector<int>(26));
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < 26; j++) {
      nxt[i][j] = INF;
      for (int k = i; k < n; k++) {
        if (s[k] == ('a' + j)) {
          nxt[i][j] = k;
          break;
        }
      }
    }
  }
  bool ans = false;
  for (int p = 1; p < (int)(t).size(); p++) {
    int dp[p + 1][(int)(t).size() - p + 1];
    for (int i = 0; i < p + 1; i++)
      for (int j = 0; j < (int)(t).size() - p + 1; j++) dp[i][j] = INF;
    dp[0][0] = -1;
    for (int i = 0; i < p + 1; i++) {
      for (int j = 0; j < (int)(t).size() - p + 1; j++) {
        int current = dp[i][j];
        if (current == INF) continue;
        if (i != p) {
          int next = nxt[current + 1][t[i] - 'a'];
          dp[i + 1][j] = min(dp[i + 1][j], next);
        }
        if ((j != (int)(t).size() - p)) {
          int next = nxt[current + 1][t[p + j] - 'a'];
          dp[i][j + 1] = min(dp[i][j + 1], next);
        }
      }
    }
    if (dp[p][(int)(t).size() - p] != INF) {
      ans = true;
      break;
    }
  }
  if ((int)(t).size() == 1) {
    for (int i = 0; i < (int)(s).size(); i++) {
      if (s[i] == t[0]) {
        ans = true;
        break;
      }
    }
  }
  cout << (ans ? "YES" : "NO") << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) solve();
  return 0;
}
