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("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
void solve() {
  string s, t;
  cin >> s;
  cin >> t;
  long long n = (long long)s.length();
  long long m = (long long)t.length();
  long long tpoint = 0;
  for (long long i = 0; i < n && tpoint < m; i++)
    if (s[i] == t[tpoint]) tpoint++;
  if (tpoint == m) {
    cout << "YES\n";
    return;
  }
  long long ok = false;
  for (long long split = 0; split < (m - 1); split++) {
    string s1 = "", s2 = "";
    for (long long i = 0; i <= split; i++) s1 += t[i];
    for (long long i = split + 1; i < m; i++) s2 += t[i];
    long long l1 = (split + 1), l2 = (m - split - 1);
    vector<vector<long long>> dp(n, vector<long long>(l1 + 1, -1));
    if (s[0] == s1[0]) {
      dp[0][1] = 0;
    }
    if (s[0] == s2[0]) {
      dp[0][0] = 1;
    } else {
      dp[0][0] = 0;
    }
    for (long long i = 1; i < n; i++) {
      for (long long j = 0; j <= min(i + 1, l1); j++) {
        dp[i][j] = dp[i - 1][j];
        if (dp[i - 1][j] == l2) continue;
        if (dp[i - 1][j] != -1 && s2[dp[i - 1][j]] == s[i])
          dp[i][j] = max(dp[i][j], 1 + dp[i - 1][j]);
        if (j > 0 && dp[i - 1][j - 1] != -1 && s1[j - 1] == s[i])
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    }
    ok |= dp[n - 1][l1] == l2;
  }
  if (ok)
    cout << "YES\n";
  else
    cout << "NO\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
