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 int N = 5e2;
string s, t;
int dp[N][N];
int is[N][28];
void solve() {
  cin >> s >> t;
  string now = "";
  for (int i = 0; i < int(s.size()); ++i) fill(is[i], is[i] + 26, 0);
  for (int i = 0; i <= int(t.size()); ++i) {
    for (int j = 0; j <= int(t.size()); ++j) dp[i][j] = 1e9;
  }
  for (int i = int(s.size()); i >= 0; --i) {
    for (int j = 0; j < 27; ++j) is[i][j] = -1;
  }
  for (int i = int(s.size()) - 1; i >= 0; --i) {
    for (int j = 0; j < 27; ++j) is[i][j] = is[i + 1][j];
    is[i][s[i] - 'a'] = i;
  }
  for (int i = 0; i < int(t.size()); ++i) {
    now += t[i];
    string rnow = "";
    for (int j = i + 1; j < int(t.size()); ++j) rnow += t[j];
    for (int j = 0; j <= int(now.size()); ++j) {
      for (int k = 0; k <= int(rnow.size()); ++k) {
        dp[j][k] = 1e9;
      }
    }
    dp[0][0] = 0;
    for (int j = 0; j <= int(now.size()); ++j) {
      for (int k = 0; k <= int(rnow.size()); ++k) {
        if (j > 0 && dp[j - 1][k] < 1e9)
          if (is[dp[j - 1][k]][now[j - 1] - 'a'] != -1)
            dp[j][k] = is[dp[j - 1][k]][now[j - 1] - 'a'] + 1;
        if (k > 0 && dp[j][k - 1] < 1e9)
          if (is[dp[j][k - 1]][rnow[k - 1] - 'a'] != -1)
            dp[j][k] = min(dp[j][k], is[dp[j][k - 1]][rnow[k - 1] - 'a'] + 1);
      }
    }
    if (dp[int(now.size())][int(rnow.size())] < 1e9) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
