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 long long N = 409;
string second, t;
long long n;
long long nx[N][26];
bool check(string a, string b) {
  a = '#' + a, b = '#' + b;
  long long x = a.size(), y = b.size();
  long long dp[x][y];
  fill(*dp, *dp + x * y, n);
  dp[0][0] = 0;
  for (long long i = 0; i < x; i++) {
    for (long long j = 0; j < y; j++) {
      if (i) dp[i][j] = min(dp[i][j], nx[dp[i - 1][j]][a[i] - 'a']);
      if (j) dp[i][j] = min(dp[i][j], nx[dp[i][j - 1]][b[j] - 'a']);
    }
  }
  if (dp[x - 1][y - 1] < n)
    return 1;
  else
    return 0;
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    cin >> second >> t;
    second = '#' + second;
    n = second.size();
    for (long long i = 0; i < 26; i++) nx[n - 1][i] = n, nx[n][i] = n;
    for (long long i = n - 2; i >= 0; i--) {
      for (long long j = 0; j < 26; j++) nx[i][j] = nx[i + 1][j];
      nx[i][second[i + 1] - 'a'] = i + 1;
    }
    long long ok = 0;
    string x = "";
    for (long long i = 0; i < t.size(); i++) {
      if (check(x, t.substr(i))) {
        ok = 1;
        break;
      }
      x += t[i];
    }
    if (ok)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
