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;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, int n) {
  stringstream s;
  s << "[";
  for (int i = 0; i < n - 1; i++) s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
inline void __anthro_begin(int line) { cerr << "#" << line << ": "; }
template <typename T>
inline void __anthro_out_var(vector<T> val) {
  cerr << arrStr(val, val.size());
}
template <typename T>
inline void __anthro_out_var(T* val) {
  cerr << arrStr(val, 2);
}
template <typename T>
inline void __anthro_out_var(T val) {
  cerr << val;
}
inline void __anthro(vector<string>::iterator it) { cerr << endl; }
template <typename T, typename... Args>
inline void __anthro(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
  __anthro_out_var(a);
  cerr << "; ";
  __anthro(++it, args...);
}
int const INF = (int)1e9 + 1e3;
long long const INFL = (long long)1e18 + 1e6;
string s, t;
vector<vector<int>> suf;
bool dpfun(string a, string b) {
  vector<vector<int>> dp(a.length() + 1, vector<int>(b.length() + 1, INF));
  dp[0][0] = 0;
  for (int i = 0; i <= a.length(); i++) {
    for (int j = 0; j <= b.length(); j++) {
      if (dp[i][j] <= s.length()) {
        int l = dp[i][j];
        if (i < a.length() && suf[l][a[i]] < INF) {
          dp[i + 1][j] = min(dp[i + 1][j], suf[l][a[i]] + 1);
        }
        if (j < b.length() && suf[l][b[j]] < INF) {
          dp[i][j + 1] = min(dp[i][j + 1], suf[l][b[j]] + 1);
        }
      }
    }
  }
  return dp[a.length()][b.length()] < INF;
}
int main() {
  int n;
  cin >> n;
  while (n--) {
    cin >> s >> t;
    for (auto& c : s) c -= 'a';
    for (auto& c : t) c -= 'a';
    suf.assign(s.length() + 1, vector<int>(26, INF));
    for (int i = s.length() - 1; i >= 0; i--) {
      suf[i] = suf[i + 1];
      suf[i][s[i]] = i;
    }
    bool flag = 0;
    for (int i = 0; i < t.length(); i++) {
      if (dpfun(t.substr(0, i), t.substr(i, t.length()))) {
        flag = 1;
        cout << "YES" << endl;
        break;
      }
    }
    if (flag == 0) cout << "NO" << endl;
  }
  return 0;
}
