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 double pi = acos(-1);
const int mod = 1000000007;
const int inf = 0x3fffffff;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
struct PairHash {
  template <typename T1, typename T2>
  std::size_t operator()(const pair<T1, T2> &p) const {
    return hash<T1>()(p.first) ^ hash<T2>()(p.second);
  }
};
string s, p;
bool delete_from(int i, int j) {
  int idx = 0;
  for (int k = i; k < j; k++) {
    while (idx < s.size() && s[idx] != p[k]) idx++;
    if (idx == s.size()) return false;
    idx++;
  }
  return true;
}
int f[401][401];
int nextc[401][26];
bool can_from(int first) {
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (int fu = 0; fu <= first; fu++) {
    for (int su = 0; su <= p.size() - first; su++) {
      if (f[fu][su] != -1) {
        if (fu + 1 <= first) {
          int ou = nextc[f[fu][su]][p[fu] - 'a'];
          if (ou != -1 && (f[fu + 1][su] == -1 || f[fu + 1][su] > ou)) {
            f[fu + 1][su] = ou;
          }
        }
        if (su + first + 1 <= p.size()) {
          int ou = nextc[f[fu][su]][p[su + first] - 'a'];
          if (ou != -1 && (f[fu][su + 1] == -1 || f[fu][su + 1] > ou)) {
            f[fu][su + 1] = ou;
          }
        }
      }
    }
  }
  return f[first][p.size() - first] != -1;
}
void solve(int ncase) {
  cin >> s >> p;
  if (s == p) {
    cout << "YES" << endl;
    return;
  }
  if (delete_from(0, p.size())) {
    cout << "YES" << endl;
    return;
  }
  vector<int> zs(26);
  vector<int> zp(26);
  for (auto c : s) zs[c - 'a']++;
  for (auto c : p) zp[c - 'a']++;
  for (int i = 0; i < 26; i++) {
    if (zs[i] < zp[i]) {
      cout << "NO" << endl;
      return;
    }
  }
  vector<int> next(26, -1);
  for (int i = s.size() - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++) nextc[i + 1][j] = next[j];
    next[s[i] - 'a'] = i + 1;
  }
  for (int j = 0; j < 26; j++) nextc[0][j] = next[j];
  for (int f = 1; f < p.size(); f++) {
    if (!delete_from(0, f) || !delete_from(f, p.size())) {
      continue;
    }
    if (can_from(f)) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
int main() {
  cout << std::fixed;
  cout << setprecision(9);
  int T = 1;
  cin >> T;
  int ncase = 0;
  while (T--) {
    solve(++ncase);
  }
}
