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;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
string s;
int dp[401][401];
bool dp2[401][401];
bool solve(string t1, string t2) {
  t1.insert(t1.begin(), '.');
  t2.insert(t2.begin(), '.');
  for (int i = 0; i < s.length(); i++) {
    for (int j = 0; j < t1.length(); j++) {
      dp[i][j] = 0;
      dp2[i][j] = 0;
    }
    dp2[i][0] = 1;
  }
  for (int i = 1; i < s.length(); i++) {
    for (int j = 0; j < t1.length(); j++) {
      if (j > 0 && t1[j] == s[i] && dp2[i - 1][j - 1]) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
        dp2[i][j] = 1;
      }
      if (t2[dp[i - 1][j] + 1] == s[i] && dp2[i - 1][j]) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1);
      }
      if (dp2[i - 1][j]) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
        dp2[i][j] = 1;
      }
    }
    if (dp[i][t1.length() - 1] == t2.length() - 1) {
      return 1;
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> s;
    s.insert(s.begin(), '.');
    string t;
    cin >> t;
    bool ans = 0;
    if (t.size() == 1) {
      for (char i : s)
        if (t[0] == i) ans = 1;
    } else {
      string t1 = "";
      string t2 = t;
      for (int i = 0; i < t.length() - 1; i++) {
        t1 += t[i];
        t2.erase(t2.begin());
        ans |= solve(t1, t2);
      }
    }
    if (ans)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
