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 = 401;
const int MOD = 1e9 + 7;
const double PI = 2 * acos(0.0);
int stop;
string s, t;
bool dp[N][N][N], vp[N][N][N];
bool fun(int pos1, int pos2, int pos3) {
  if (pos3 == s.size()) {
    return pos1 == stop && pos2 == t.size();
  }
  if (pos1 == stop && pos2 == t.size()) {
    return 1;
  }
  bool &tmp = dp[pos1][pos2][pos3];
  bool &vmp = vp[pos1][pos2][pos3];
  if (vmp) {
    return tmp;
  }
  tmp = 0, vmp = 1;
  if (pos1 < stop && t[pos1] == s[pos3]) {
    tmp |= fun(pos1 + 1, pos2, pos3 + 1);
  }
  if (pos2 < t.size() && t[pos2] == s[pos3]) {
    tmp |= fun(pos1, pos2 + 1, pos3 + 1);
  }
  tmp |= fun(pos1, pos2, pos3 + 1);
  return tmp;
}
int main() {
  int test;
  scanf("%d", &test);
  while (test--) {
    bool ok = 0;
    cin >> s >> t;
    memset(vp, 0, sizeof vp);
    for (int i = 0; i < t.size() - 1; i++) {
      stop = i + 1;
      if (fun(0, i + 1, 0)) {
        ok = 1;
        break;
      }
    }
    stop = 0;
    ok |= fun(0, 0, 0);
    stop = t.size();
    ok |= fun(0, t.size(), 0);
    puts(ok ? "YES" : "NO");
  }
  return 0;
}
