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 mod = 1000000007;
const vector<int> mods = {998244353, 1000000007, 1000000009};
const long long inf = 3e18;
const double pi = acos(-1.0);
const double eps = 0.0000001;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class... K>
using umap = unordered_map<K...>;
template <class... K>
using uset = unordered_set<K...>;
long long T;
string s, t;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.precision(12);
  cin >> T;
  for (long long _T = (0); _T < (T); ++_T) {
    cin >> s >> t;
    long long sn = s.length(), tn = t.length();
    for (long long st = (0); st < (tn + 1); ++st) {
      vector<long long> dp(st + 1, -1);
      dp[0] = 0;
      for (long long i = (0); i < (sn); ++i) {
        if (st == 1) {
        }
        vector<long long> nxdp = dp;
        for (long long j = (0); j < (st + 1); ++j) {
          nxdp[j] = max(nxdp[j], dp[j]);
          if (j < st && s[i] == t[j]) {
            nxdp[j + 1] = max(nxdp[j + 1], dp[j]);
          }
          if (dp[j] >= 0 && s[i] == t[dp[j] + st]) {
            nxdp[j] = max(nxdp[j], dp[j] + 1);
          }
        }
        dp = nxdp;
      }
      if (dp[st] == tn - st) {
        cout << "YES\n";
        goto outer;
      }
    }
    cout << "NO\n";
  outer:;
  }
  return 0;
}
