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 long long N = 100009;
vector<pair<int, int> > tmp;
set<pair<int, int> > myset;
int n, m;
long long ok[409][409], run;
string s, t;
void IO() {
  int test;
  cin >> test;
  while (test--) {
    cin >> s;
    n = s.size();
    s = '#' + s;
    cin >> t;
    m = t.size();
    t = '#' + t;
    run = 0;
    memset(ok, 0, sizeof(ok));
    int done = 0;
    for (int ss = 1; ss <= m; ++ss) {
      for (int i = 0; i <= m; ++i)
        for (int j = 0; j <= m; ++j) ok[i][j] = 0;
      myset.clear();
      myset.insert(make_pair(0, 1));
      myset.insert(make_pair(1, 0));
      int nn = m - ss;
      ++run;
      ok[0][0] = run;
      for (int i = 1; i <= n; ++i) {
        tmp.clear();
        for (auto x : myset) {
          int j = x.first, k = x.second;
          if (j > 0 && ok[j - 1][k] == run && s[i] == t[j] ||
              k > 0 && ok[j][k - 1] == run && s[i] == t[ss + k])
            ok[j][k] = run, tmp.push_back(make_pair(j, k));
        }
        for (auto x : tmp) {
          myset.erase(x);
          if (x.first + 1 <= ss) myset.insert(make_pair(x.first + 1, x.second));
          if (x.second + 1 <= nn)
            myset.insert(make_pair(x.first, x.second + 1));
        }
      }
      if (ok[ss][nn] == run) {
        done = 1;
        break;
      }
    }
    if (done)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
}
int main() {
  IO();
  return 0;
}
