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 = 405;
int dp[N][N];
string s, a, b;
bool check() {
  for (int i = 0; i <= a.length(); ++i)
    for (int j = 0; j <= b.length(); ++j) dp[i][j] = 0;
  for (int i = 0; i <= a.length(); ++i)
    for (int j = i == 0; j <= b.length(); ++j) {
      dp[i][j] = INT_MAX;
      if (i - 1 >= 0) {
        int pos = dp[i - 1][j] + 1;
        while (pos <= s.length() && s[pos - 1] != a[i - 1]) ++pos;
        if (pos <= s.length()) dp[i][j] = min(dp[i][j], pos);
      }
      if (j - 1 >= 0) {
        int pos = dp[i][j - 1] + 1;
        while (pos <= s.length() && s[pos - 1] != b[j - 1]) ++pos;
        if (pos <= s.length()) dp[i][j] = min(dp[i][j], pos);
      }
    }
  return dp[a.length()][b.length()] != INT_MAX;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    bool ok = 0;
    cin >> s >> b;
    a.clear();
    while (b.length()) {
      ok |= check();
      a += b.front();
      b.erase(0, 1);
    }
    puts(ok ? "YES" : "NO");
  }
  return 0;
}
