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 N = 405;
const int inf = 1e9;
int dp[N][N], n, m, nxt[N][27];
string s, t;
bool check(const string &u, const string &v) {
  string a = u;
  string b = v;
  int sza = a.size();
  int szb = b.size();
  a = " " + a;
  b = " " + b;
  for (int i = 0; i <= sza; i++) {
    for (int j = 0; j <= szb; j++) dp[i][j] = inf;
  }
  dp[0][0] = 0;
  for (int i = 0; i <= sza; i++) {
    for (int j = 0; j <= szb; j++) {
      if (i == 0 && j == 0) continue;
      if (j && dp[i][j - 1] != 1e9)
        dp[i][j] = min(dp[i][j], nxt[dp[i][j - 1]][b[j] - 'a']);
      if (i && dp[i - 1][j] != 1e9)
        dp[i][j] = min(dp[i][j], nxt[dp[i - 1][j]][a[i] - 'a']);
    }
  }
  return (dp[sza][szb] != inf);
}
void sol() {
  cin >> s >> t;
  n = s.size();
  m = t.size();
  s = " " + s;
  t = " " + t;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= 'z' - 'a'; j++) {
      nxt[i][j] = inf;
      for (int k = i + 1; k <= n; k++) {
        if (s[k] - 'a' == j) {
          nxt[i][j] = k;
          break;
        }
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    if (check(t.substr(1, i), t.substr(i + 1, m - i))) {
      cout << "YES" << '\n';
      return;
    }
  }
  cout << "NO" << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) sol();
}
