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 nxt[405][26];
int dp[405][405];
bool foo(string& s, string& a, string b) {
  int i, j;
  for (i = 0; i < (int)((a).size()) + 1; i++) {
    for (j = 0; j < (int)((b).size()) + 1; j++) dp[i][j] = 1e9;
  }
  dp[0][0] = 0;
  for (i = 0; i < (int)((a).size()) + 1; i++) {
    for (j = 0; j < (int)((b).size()) + 1; j++) {
      int len = dp[i][j];
      if (len == 1e9) continue;
      if (i < (int)((a).size())) {
        int pos = nxt[len + 1][a[i] - 'a'];
        if (pos != -1) {
          dp[i + 1][j] = min(dp[i + 1][j], pos);
        }
      }
      if (j < (int)((b).size())) {
        int pos = nxt[len + 1][b[j] - 'a'];
        if (pos != -1) {
          dp[i][j + 1] = min(dp[i][j + 1], pos);
        }
      }
    }
  }
  if (dp[(int)((a).size())][(int)((b).size())] == 1e9) return 0;
  return 1;
}
int main() {
  int n, i, j, t;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    string s, t;
    cin >> s >> t;
    for (i = 0; i < 26; i++) nxt[(int)((s).size()) + 1][i] = -1;
    for (i = (int)((s).size()); i >= 1; i--) {
      for (j = 0; j < 26; j++) {
        nxt[i][j] = nxt[i + 1][j];
        if (s[i - 1] == 'a' + j) nxt[i][j] = i;
      }
    }
    string a;
    bool ans = 0;
    for (i = 0; i < (int)((t).size()); i++) {
      ans = foo(s, a, t.substr(i, (int)((t).size()) - i));
      if (ans) break;
      a.push_back(t[i]);
    }
    if (ans)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
