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 mod = 1e9 + 7;
const int maxn = 4e2 + 10;
const int inf = 0x3f3f3f3f;
string s, str, a, b;
int nex[maxn][27], dp[maxn][maxn];
bool check() {
  memset(dp, inf, sizeof(dp));
  dp[0][0] = -1;
  for (int i = (0); i <= (a.size()); i++) {
    for (int j = (0); j <= (b.size()); j++) {
      if (i == a.size() && j == b.size()) continue;
      if (dp[i][j] == inf) continue;
      if (i < a.size()) {
        dp[i + 1][j] = min(dp[i + 1][j], nex[dp[i][j] + 1][a[i] - 'a']);
      }
      if (j < b.size()) {
        dp[i][j + 1] = min(dp[i][j + 1], nex[dp[i][j] + 1][b[j] - 'a']);
      }
    }
  }
  if (dp[a.size()][b.size()] != inf)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> s >> str;
    memset(nex, inf, sizeof(nex));
    for (int i = (s.size() - 1); i >= (0); i--) {
      for (int j = (0); j <= (26); j++) nex[i][j] = nex[i + 1][j];
      nex[i][s[i] - 'a'] = i;
    }
    bool f = false;
    for (int i = (1); i <= (str.size()); i++) {
      a = str.substr(0, i);
      b = str.substr(i, str.size() - i);
      if (check()) {
        f = true;
        break;
      }
    }
    cout << (f ? "YES" : "NO") << '\n';
  }
  return 0;
}
