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>
const int N = 1e5 + 5;
using namespace std;
int T;
int F[500];
int Pre[500];
void nhap() { cin >> T; }
void xuli() {
  while (T--) {
    string S, T;
    cin >> S >> T;
    int t;
    t = 0;
    for (int i = 0; i < S.size(); i++)
      if (t < T.size() && S[i] == T[t]) t++;
    if (t == T.size()) {
      cout << "YES\n";
      continue;
    }
    int cc = 0;
    for (int i = 0; i < T.size(); i++) {
      string A = T.substr(0, i);
      string B = T.substr(i);
      F[0] = 0;
      Pre[0] = 0;
      for (int j = 1; j <= A.size(); j++) F[j] = -1, Pre[j] = -1;
      for (int j = 0; j < S.size(); j++) {
        for (int k = 0; k < A.size(); k++)
          if (S[j] == A[k]) F[k + 1] = max(Pre[k + 1], Pre[k]);
        for (int k = 0; k <= A.size(); k++)
          if (S[j] == B[Pre[k]]) F[k] = max(F[k], Pre[k] + 1);
        for (int k = 0; k <= A.size(); k++) Pre[k] = F[k];
      }
      if (F[A.size()] == B.size()) {
        cc = 1;
        cout << "YES\n";
        break;
      }
    }
    if (!cc) cout << "NO\n";
  }
}
int main() {
  nhap();
  xuli();
}
