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 NN = 1e6 + 7;
const int N = 1e3 + 7;
const int M = 20;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 7;
const double long rf = 1e-14;
string s, t;
int d[N][N];
int p[N];
int nxt[N][26];
bool foo(int la, int lb) {
  for (int i = 0; i <= la; i++) {
    for (int j = 0; j <= lb; j++) {
      d[i][j] = 1e9;
    }
  }
  d[0][0] = -1;
  for (int i = 0; i <= la; i++) {
    for (int j = 0; j <= lb; j++) {
      if (d[i][j] != 1e9) {
        if (j < lb)
          d[i][j + 1] = min(d[i][j + 1], nxt[d[i][j] + 1][t[j + la] - 'a']);
        if (i < la)
          d[i + 1][j] = min(d[i + 1][j], nxt[d[i][j] + 1][t[i] - 'a']);
      }
    }
  }
  if (d[la][lb] == 1e9) return false;
  return true;
}
void solve1() {
  cin >> s >> t;
  int n1 = (int)s.size();
  int n2 = (int)t.size();
  for (int i = 0; i <= n1 + 10; i++) {
    for (int j = 0; j < 26; j++) {
      nxt[i][j] = 1e9;
      p[j] = 1e9;
    }
  }
  for (int i = n1 - 1; i >= 0; i--) {
    p[s[i] - 'a'] = i;
    for (int j = 0; j < 26; j++) {
      nxt[i][j] = p[j];
    }
  }
  for (int i = 0; i < n2; i++) {
    if (foo(i, n2 - i)) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  int cghf = 1;
  cin >> cghf;
  while (cghf--) {
    solve1();
  }
}
