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 string __sep = "\n";
struct triple {
  int x, y, z;
  triple(int _x, int _y, int _z) : x(_x), y(_y), z(_z) {}
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int Q = 1;
  cin >> Q;
  while (Q-- > 0) {
    string s, t;
    cin >> s >> t;
    bool success = false;
    for (int sep = 1; sep <= t.length(); ++sep) {
      string first = t.substr(0, sep);
      string sec = "";
      if (sep < t.length()) sec = t.substr(sep, t.length() - sep);
      int cur = 0;
      int n = s.length();
      bool fail = false;
      int d[n + 1][first.length() + 1];
      const int INF = 0;
      for (int i = 0; i <= n; ++i) {
        for (int j = 0; j <= first.length(); ++j) d[i][j] = -1;
      }
      d[0][0] = 0;
      for (int i = 1; i <= n; ++i) {
        char cur = s[i - 1];
        for (int j = 0; j <= i && j <= first.length(); ++j) {
          d[i][j] = d[i - 1][j];
          if (j > 0 && first[j - 1] == cur) {
            d[i][j] = max(d[i][j], d[i - 1][j - 1]);
          }
          if (d[i - 1][j] >= 0 && d[i - 1][j] < sec.length() &&
              sec[d[i - 1][j]] == cur) {
            d[i][j] = max(d[i][j], 1 + d[i - 1][j]);
          }
        }
      }
      for (int i = 0; i <= n; ++i) {
        if (d[i][first.length()] >= (int)sec.length()) {
          success = true;
          break;
        }
      }
      if (success) break;
    }
    if (success)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
