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;
int dp[409][409];
string s, t1, t2, t;
int n, m1, m2, m;
int go(int i, int j) {
  if (i == -1 && j == 0) return 0;
  if (i == -1) return -2;
  if (dp[i][j] != -1) return dp[i][j];
  int ans1 = go(i - 1, j);
  if (ans1 != -2 && ans1 < t2.size()) {
    if (s[i] == t2[ans1]) ans1++;
  }
  int ans2 = -2;
  if (j > 0 && s[i] == t1[j - 1]) ans2 = go(i - 1, j - 1);
  return dp[i][j] = max(ans1, ans2);
}
int v(int x) {
  t1 = "";
  t2 = "";
  for (int i = 0; i <= x; i++) t1 += t[i];
  for (int i = x + 1; i < m; i++) t2 += t[i];
  m1 = t1.size();
  m2 = t2.size();
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = -1;
  return go(n - 1, m1);
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    cin >> s >> t;
    n = s.size(), m = t.size();
    bool bl = false;
    for (int i = 0; i < m; i++) {
      if (v(i) == m - i - 1) {
        bl = true;
        break;
      }
    }
    if (bl)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
