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;
template <typename T>
void setmax(T& x, T y) {
  x = max(x, y);
}
template <typename T>
void setmin(T& x, T y) {
  x = min(x, y);
}
const double PI = acos(-1.0);
const long long INF = 1e9 + 47;
const long long LINF = INF * INF;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 400 + 7;
int dp[N][N];
inline bool go(string& a, string& l, string& r) {
  for (int i = (0); i < ((int)((a).size()) + 1); ++i)
    for (int j = (0); j < ((int)((l).size()) + 1); ++j) dp[i][j] = -INF;
  dp[0][0] = 0;
  for (int i = (0); i < ((int)((a).size())); ++i)
    for (int j = (0); j < ((int)((l).size()) + 1); ++j) {
      if (dp[i][j] < 0) continue;
      if (a[i] == l[j]) setmax(dp[i + 1][j + 1], dp[i][j]);
      if (dp[i][j] < (int)((r).size()) && a[i] == r[dp[i][j]])
        setmax(dp[i + 1][j], dp[i][j] + 1);
      setmax(dp[i + 1][j], dp[i][j]);
    }
  return dp[(int)((a).size())][(int)((l).size())] == (int)((r).size());
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string a, b;
    cin >> a >> b;
    bool ok = false;
    for (int i = 0; i < (int)((b).size()) && !ok; ++i) {
      string l, r;
      for (int j = (0); j < (i + 1); ++j) l += b[j];
      for (int j = (i + 1); j < ((int)((b).size())); ++j) r += b[j];
      ok |= go(a, l, r);
    }
    cout << (ok ? "YES\n" : "NO\n");
  }
  cerr << "Time elapsed: " << clock() / (double)CLOCKS_PER_SEC << endl;
  return 0;
}
