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 = 410, maxl = 26, inf = 1e9;
int dp[maxn][maxn], nxt[maxn][maxl];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    string a, t;
    cin >> a >> t;
    for (int i = 0; i < (int)a.size(); i++)
      for (int j = 0; j < maxl; j++) {
        nxt[i][j] = -1;
        for (int k = i; k < (int)a.size(); k++)
          if (a[k] == j + 'a') {
            nxt[i][j] = k;
            break;
          }
      }
    bool find = false;
    for (int x = 0; x < (int)t.size(); x++) {
      string b = "", c = "";
      for (int i = 0; i < (int)t.size(); i++)
        if (i < x)
          b += t[i];
        else
          c += t[i];
      for (int i = 0; i <= (int)b.size(); i++)
        for (int j = 0; j <= (int)c.size(); j++) {
          dp[i][j] = (i + j == 0 ? 0 : inf);
          if (i > 0 && dp[i - 1][j] < (int)a.size() &&
              nxt[dp[i - 1][j]][b[i - 1] - 'a'] != -1)
            dp[i][j] = min(dp[i][j], 1 + nxt[dp[i - 1][j]][b[i - 1] - 'a']);
          if (j > 0 && dp[i][j - 1] < (int)a.size() &&
              nxt[dp[i][j - 1]][c[j - 1] - 'a'] != -1)
            dp[i][j] = min(dp[i][j], 1 + nxt[dp[i][j - 1]][c[j - 1] - 'a']);
        }
      if (dp[(int)b.size()][(int)c.size()] <= (int)a.size()) {
        find = true;
        break;
      }
    }
    cout << (find ? "YES\n" : "NO\n");
  }
  return 0;
}
