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 maxn = 405;
const int inf = 0x3f3f3f3f;
int dp[maxn][maxn];
bool check(string s, string t1, string t2) {
  memset(dp, -1, sizeof(dp));
  int slen = s.size();
  int t1len = t1.size();
  dp[0][0] = 0;
  for (int i = 0; i < slen; i++) {
    for (int j = 0; j <= t1len && j <= i; j++) {
      if (dp[i][j] != -1) {
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
        if (s[i] == t1[j]) {
          dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
        }
        if (s[i] == t2[dp[i][j]]) {
          dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + 1);
        }
      }
    }
  }
  for (int i = t1len; i <= slen; i++) {
    if (dp[i][t1len] == t2.size()) return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    string s, t;
    cin >> s >> t;
    bool f = false;
    for (int i = 0; i <= t.size(); i++) {
      string t1 = t.substr(0, i);
      string t2 = t.substr(i, t.size() - i);
      if (check(s, t1, t2)) {
        f = true;
        break;
      }
    }
    if (f) {
      puts("YES");
    } else {
      puts("NO");
    }
  }
  return 0;
}
