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 dydis = 401;
const int inf = 1e9;
int dp[dydis][dydis] = {0};
bool tinka(string visas, string a, string b) {
  for (int i = 0; i <= visas.size(); i++) {
    for (int j = 0; j <= a.size(); j++) {
      dp[i][j] = -inf;
    }
  }
  dp[0][0] = 0;
  for (int i = 0; i < visas.size(); i++) {
    for (int j = 0; j <= a.size(); j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (dp[i][j] < b.size() && visas[i] == b[dp[i][j]]) {
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + 1);
      }
      if (j != a.size() && a[j] == visas[i]) {
        dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      }
    }
  }
  return dp[visas.size()][a.size()] == b.size();
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    string a, b;
    cin >> a >> b;
    string ans = "NO";
    for (int i = 0; i < b.size(); i++) {
      string kair = "", desn = "";
      if (i != 0) kair = b.substr(0, i);
      desn = b.substr(i, b.size() - i);
      if (tinka(a, kair, desn)) ans = "YES";
    }
    cout << ans << "\n";
  }
  return 0;
}
