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 = 1e5 + 10, M = 2e5 + 10, M2 = 1e6 + 10, mod = 1e9 + 7,
          inf = 1e17 + 10;
const int INF = 1e9 + 7;
void add(int& a, int b) {
  a += b;
  if (a >= mod) {
    a -= mod;
  }
}
string s;
string t;
int n, m;
int cache[401][401][401];
int dp(int i, int j, int k) {
  if (i == n) {
    if (k == m)
      return 0;
    else
      return -1e10;
  }
  if (j == m && k == m) return 0;
  if (j >= k) return -1e10;
  int& ans = cache[i][j][k];
  if (ans != -1) return ans;
  ans = dp(i + 1, j, k);
  if (i < n && s[i] == t[j]) ans = max(ans, dp(i + 1, j + 1, k) + 1);
  if (j < n && s[i] == t[k]) ans = max(ans, dp(i + 1, j, k + 1) + 1);
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    cin >> s >> t;
    n = s.size();
    m = t.size();
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= n; j++) {
        for (int k = 0; k <= n; k++) {
          cache[i][j][k] = -1;
        }
      }
    }
    bool ok = false;
    for (int i = 0; i <= n; i++) {
      if (dp(0, 0, i) >= m) ok = true;
    }
    if (ok)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
