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;
const int INF = 2e9;
const long long INFLL = 4e18L;
const int MOD = 0;
const int N = 400 + 3;
int dp[N][N];
int nxt[N][26];
bool solve(string& s, string& a, string& b) {
  for (int i = 0; i <= (int)a.size(); ++i)
    for (int j = 0; j <= (int)b.size(); ++j) dp[i][j] = INF;
  dp[0][0] = 0;
  for (int i = 0; i <= (int)a.size(); ++i)
    for (int j = 0; j <= (int)b.size(); ++j) {
      int cur_s = dp[i][j];
      if (cur_s > (int)s.size()) continue;
      if (i < (int)a.size()) {
        int nxt_a = nxt[cur_s][a[i] - 'a'];
        if (nxt_a < INF) dp[i + 1][j] = min(dp[i + 1][j], nxt_a + 1);
      }
      if (j < (int)b.size()) {
        int nxt_b = nxt[cur_s][b[j] - 'a'];
        if (nxt_b < INF) dp[i][j + 1] = min(dp[i][j + 1], nxt_b + 1);
      }
    }
  return dp[(int)a.size()][(int)b.size()] < INF;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(16);
  int tc;
  cin >> tc;
  while (tc--) {
    string s, t;
    cin >> s >> t;
    for (int c = 0; c <= 25; ++c) nxt[(int)s.size()][c] = INF;
    for (int i = (int)s.size() - 1; i >= 0; --i) {
      for (int c = 0; c <= 25; ++c) nxt[i][c] = nxt[i + 1][c];
      nxt[i][s[i] - 'a'] = i;
    }
    bool possible = false;
    for (int nb = 1; nb <= (int)t.size(); ++nb) {
      string s1 = t.substr(0, nb);
      string s2 = t.substr(nb);
      possible |= solve(s, s1, s2);
    }
    cout << (possible ? "YES\n" : "NO\n");
  }
}
