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;
template <typename T, typename V>
ostream &operator<<(ostream &os, const pair<T, V> pai) {
  return os << '(' << pai.first << ' ' << pai.second << ')';
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> v) {
  cout << '[';
  for (auto p : v) cout << p << ",";
  cout << "]";
  return os;
}
template <typename T, typename V>
ostream &operator<<(ostream &os, const set<T, V> v) {
  cout << "{";
  for (auto p : v) cout << p << ",";
  cout << "}";
  return os;
}
template <typename T, typename V>
ostream &operator<<(ostream &os, const map<T, V> v) {
  cout << "{";
  for (auto p : v) cout << p << ",";
  cout << "}";
  return os;
}
template <typename T>
void get_max(T &a, T b) {
  a = max(a, b);
}
template <typename T>
void get_min(T &a, T b) {
  a = min(a, b);
}
const int MAXN = 1e5 + 10;
struct X {
  int s, a, b;
};
char CK[410][410][410];
int T;
string x, y;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> T;
  for (int t = 1; t <= T; t++) {
    string x, y;
    cin >> x >> y;
    vector<X> V;
    for (int i = 1; i <= y.size(); i++) V.push_back({i, 0, i});
    for (int i = 0; i < x.size(); i++) {
      vector<X> NV;
      for (auto val : V) {
        int a = val.a, b = val.b;
        int ck = 0;
        if (a < val.s && x[i] == y[a]) {
          ck = 1;
          if (CK[val.s][a + 1][b] != t) {
            CK[val.s][a + 1][b] = t;
            NV.push_back({val.s, a + 1, b});
          }
        }
        if (b < y.size() && x[i] == y[b]) {
          ck = 1;
          if (CK[val.s][a][b + 1] != t) {
            CK[val.s][a][b + 1] = t;
            NV.push_back({val.s, a, b + 1});
          }
        }
        if (ck == 0) NV.push_back({val.s, a, b});
      }
      V = NV;
    }
    int ck = 0;
    for (int i = 1; i <= y.size(); i++) {
      if (CK[i][i][y.size()] == t) ck = 1;
    }
    if (ck)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
