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>
#pragma GCC optimize("O3")
using namespace std;
string to_string(const string &s) { return '"' + s + '"'; }
void debug_out() { cout << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug_out(T...);
}
using ll = int_fast64_t;
using ld = long double;
const int N = 4e2 + 5, C = 26, INF = 1e9 + 7;
int nxt[N][C], dp[N][N];
void solve() {
  string s, t;
  cin >> s >> t;
  for (auto &c : s) c -= 'a';
  for (auto &c : t) c -= 'a';
  int ns = s.length(), nt = t.length();
  fill(&nxt[0][0], &nxt[0][0] + sizeof(nxt) / sizeof(nxt[0][0]), INF);
  for (int i = ns - 1; i >= 0; --i) {
    for (int j = 0; j < C; ++j) nxt[i][j] = nxt[i + 1][j];
    nxt[i][s[i]] = i;
  }
  for (int a = 0; a < nt; ++a) {
    int b = nt - a;
    fill(&dp[0][0], &dp[0][0] + sizeof(dp) / sizeof(dp[0][0]), INF);
    dp[0][0] = 0;
    for (int i = 0; i <= a; ++i) {
      for (int j = 0; j <= b; ++j) {
        int l = dp[i][j];
        if (l == INF) continue;
        if (i < a && nxt[l][t[i]] < INF) {
          dp[i + 1][j] = min(dp[i + 1][j], nxt[l][t[i]] + 1);
        }
        if (j < b && nxt[l][t[a + j]] < INF) {
          dp[i][j + 1] = min(dp[i][j + 1], nxt[l][t[a + j]] + 1);
        }
      }
    }
    if (dp[a][b] < INF) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  cin.exceptions(cin.failbit);
  ;
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    solve();
  }
}
