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;
int D[405][405];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    string C, t;
    cin >> C >> t;
    vector<vector<int>> next(512, vector<int>(C.size() + 2, C.size()));
    for (int k = 'a'; k <= 'z'; k++) {
      for (int i = C.size() - 1; i >= 0; i--) {
        if (C[i] == k)
          next[k][i] = i;
        else
          next[k][i] = next[k][i + 1];
      }
    }
    int flag = 0;
    for (int I = 0; I < t.size() && !flag; I++) {
      string A = " " + t.substr(0, I);
      int N = A.size() - 1;
      string B = " " + t.substr(I);
      int M = B.size() - 1;
      for (int i = 0; i <= N; i++)
        for (int j = 0; j <= M; j++) D[i][j] = C.size();
      D[0][0] = -1;
      for (int i = 1; i <= N; i++) D[i][0] = next[A[i]][D[i - 1][0] + 1];
      for (int j = 1; j <= M; j++) D[0][j] = next[B[j]][D[0][j - 1] + 1];
      for (int i = 1; i <= N; i++)
        for (int j = 1; j <= M; j++)
          D[i][j] =
              min(next[A[i]][D[i - 1][j] + 1], next[B[j]][D[i][j - 1] + 1]);
      if (D[N][M] < C.size()) flag = 1;
    }
    puts(flag ? "YES" : "NO");
  }
  return 0;
}
