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 maxN = 405;
int dp[maxN][maxN];
string s;
string t;
const int INF = (int)1e9;
int nxt[maxN][26];
void solve() {
  cin >> s >> t;
  for (int i = s.size() - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++) {
      nxt[i][j] = nxt[i + 1][j];
      if (i == s.size() - 1) nxt[i][j] = s.size() + 10;
      if (s[i] == j + 'a') nxt[i][j] = i;
    }
  }
  for (int len = 1; len <= t.size(); len++) {
    int n = t.size();
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= n; j++) {
        dp[i][j] = INF;
      }
    }
    dp[0][0] = 0;
    for (int i = 0; i <= len; i++) {
      for (int j = 0; j <= n - len; j++) {
        if (i && dp[i - 1][j] < s.size()) {
          dp[i][j] = min(nxt[dp[i - 1][j]][t[i - 1] - 'a'] + 1, dp[i][j]);
        }
        if (j && dp[i][j - 1] < s.size()) {
          dp[i][j] = min(nxt[dp[i][j - 1]][t[len + j - 1] - 'a'] + 1, dp[i][j]);
        }
      }
    }
    if (dp[len][n - len] <= s.size() + 2) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  int tst;
  cin >> tst;
  while (tst--) {
    solve();
  }
  return 0;
}
