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;
ostream_iterator<int> output(cout, " ");
long long max(long long a, long long b) { return a > b ? a : b; }
long long power(long long a, long long b) {
  long long tmp = 1LL;
  while (b) {
    if (b & 1) {
      tmp = (tmp * a);
    }
    a = (a * a);
    b /= 2;
  }
  return tmp;
}
string s, t;
int dp[407][407];
int nxt[407][27];
int n, m;
bool answer(string first, string second) {
  int f = (int)first.size(), se = (int)second.size();
  dp[0][0] = 0;
  for (int i = 0; i <= f; i++) {
    for (int j = 0; j <= se; j++) {
      if (i == 0 && j == 0) continue;
      dp[i][j] = n + 1;
      if (i != 0)
        dp[i][j] = min(dp[i][j], nxt[dp[i - 1][j]][first[i - 1] - 'a'] + 1);
      if (j != 0) {
        dp[i][j] = min(dp[i][j], nxt[dp[i][j - 1]][second[j - 1] - 'a'] + 1);
      }
    }
  }
  if (dp[f][se] <= n) return true;
  return false;
}
void solve() {
  cin >> s >> t;
  n = (int)s.size();
  m = (int)t.size();
  for (int i = n; i <= 406; i++) {
    for (int j = 0; j < 26; j++) nxt[i][j] = i + 1;
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++) {
      nxt[i][j] = nxt[i + 1][j];
    }
    nxt[i][s[i] - 'a'] = i;
  }
  for (int i = 0; i < m; i++) {
    if (answer(t.substr(0, i), t.substr(i))) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int t1;
  cin >> t1;
  while (t1--) solve();
}
