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>
#pragma GCC optimize("-Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")
using namespace std;
using ll = long long;
using ld = long double;
const int N = 400 + 7, mod = 998244353;
const int inf = 1e5;
int n, m;
string s, t;
int dp[N][N];
int solve(int i, int r1, int k) {
  if (i == n) return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    n = (int)s.size();
    m = (int)t.size();
    memset(dp, -1, sizeof dp);
    bool ok = 0;
    for (int j = 0; j <= m; ++j) {
      for (int i = 0; i < n; ++i) {
        for (int k = 0; k <= m; ++k) {
          dp[i][0] = j;
        }
      }
      for (int i = 0; i < n; ++i) {
        for (int k = 0; k <= j; ++k) {
          if (~dp[i][k]) {
            if (k < j && s[i] == t[k]) {
              dp[i + 1][k + 1] = max(dp[i + 1][k + 1], dp[i][k]);
            }
            dp[i + 1][k] = max(dp[i][k], dp[i + 1][k]);
            if (dp[i][k] < m && s[i] == t[dp[i][k]]) {
              dp[i + 1][k] = max(dp[i + 1][k], dp[i][k] + 1);
            }
          }
        }
      }
      ok |= (dp[n][j] == m);
    }
    cout << (ok ? "YES" : "NO") << "\n";
  }
  return 0;
}
