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 N = 400 + 5;
const int INF = 0x3f3f3f3f;
const long long INF_LL = 0x3f3f3f3f3f3f3f3f;
int f[N][N];
char s[N], t[N];
int n, m;
bool DP(int pos) {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) f[i][j] = -INF;
  f[0][0] = 0;
  int t1 = pos - 1;
  int t2 = m - t1;
  for (int i = 1; i <= n; i++) {
    memcpy(f[i], f[i - 1], sizeof(f[i]));
    for (int j = 0; j <= min(i, t1); j++) {
      if (j && s[i] == t[j]) f[i][j] = max(f[i][j], f[i - 1][j - 1]);
      if (f[i - 1][j] >= 0 && f[i - 1][j] < t2 && s[i] == t[pos + f[i - 1][j]])
        f[i][j] = max(f[i][j], f[i - 1][j] + 1);
    }
  }
  return f[n][t1] == t2;
}
bool Solve() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  for (int i = 1; i <= m; i++)
    if (DP(i)) return 1;
  return 0;
}
int main() {
  int t;
  cin >> t;
  while (t--) cout << (Solve() ? "YES\n" : "NO\n");
  return 0;
}
