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 = 1e9;
bool solve(string s, string t) {
  int ind = 0;
  for (int i = 0; i < (int)(s).size(); ++i) {
    if (s[i] == t[ind]) {
      ++ind;
      if (ind == (int)(t).size()) return true;
    }
  }
  vector<vector<int> > dp((int)(s).size() + 1,
                          vector<int>((int)(s).size() + 1));
  for (int i = 1; i < (int)(t).size(); ++i) {
    string pref, suf;
    for (int j = 0; j < i; ++j) pref += t[j];
    for (int j = i; j < (int)(t).size(); ++j) suf += t[j];
    for (auto &x : dp)
      for (auto &j : x) j = -inf;
    dp[0][0] = 0;
    for (int j = 1; j <= (int)(s).size(); ++j) {
      for (int have = 0; have <= i; ++have) {
        dp[j][have] = max(dp[j - 1][have], dp[j][have]);
        if (dp[j - 1][have] < (int)(suf).size() && dp[j - 1][have] >= 0 &&
            suf[dp[j - 1][have]] == s[j - 1])
          dp[j][have] = max(dp[j][have], dp[j - 1][have] + 1);
        if (have < (int)(pref).size() && pref[have] == s[j - 1])
          dp[j][have + 1] = max(dp[j][have + 1], dp[j - 1][have]);
      }
    }
    if (dp[(int)(s).size()][(int)(pref).size()] == (int)(suf).size())
      return true;
  }
  return false;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s, t;
    cin >> s >> t;
    if (solve(s, t))
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
