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 auto start_time = std::chrono::high_resolution_clock::now();
void timecalculater() {}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void __print(int x) { cout << x; }
void __print(long x) { cout << x; }
void __print(long long x) { cout << x; }
void __print(unsigned x) { cout << x; }
void __print(unsigned long x) { cout << x; }
void __print(unsigned long long x) { cout << x; }
void __print(float x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
long long p = 1e9 + 7;
const int N = 2e5 + 5;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  ;
  int T = 1;
  cin >> T;
  while (T--) {
    string s, t;
    cin >> s >> t;
    int n, m;
    string ans = "NO";
    n = (int)(s).size();
    m = (int)(t).size();
    int nxt[n + 2][26];
    for (int i = 0; i < n; i++) {
      for (int k = 0; k < 26; k++) {
        nxt[i][k] = n + 1;
        for (int j = i; j < n; j++) {
          if (s[j] - 'a' == k) {
            nxt[i][k] = j + 1;
            break;
          }
        }
      }
    }
    for (int i = 0; i < (26); ++i) nxt[n][i] = n + 1;
    for (int i = 0; i < (26); ++i) nxt[n + 1][i] = n + 1;
    for (int i = 0; i <= m; i++) {
      string a = t.substr(0, i);
      string b = t.substr(i);
      int dp[(int)(a).size() + 1][(int)(b).size() + 1];
      memset(dp, 0, sizeof(dp));
      dp[0][0] = 0;
      ;
      for (int j = 0; j <= (int)(a).size(); j++) {
        for (int k = 0; k <= (int)(b).size(); k++) {
          if (j == 0 && k == 0) continue;
          if (j == 0) {
            int u = dp[j][k - 1];
            dp[j][k] = nxt[u][b[k - 1] - 'a'];
          } else if (k == 0) {
            int u = dp[j - 1][k];
            dp[j][k] = nxt[u][a[j - 1] - 'a'];
          } else {
            int u = dp[j][k - 1];
            int u1 = dp[j - 1][k];
            dp[j][k] = min(nxt[u][b[k - 1] - 'a'], nxt[u1][a[j - 1] - 'a']);
          }
        }
      }
      if (dp[(int)(a).size()][(int)(b).size()] <= n) {
        ans = "YES";
        break;
      }
    }
    cout << ans << '\n';
  }
  timecalculater();
  return 0;
}
