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;
int dp[maxn][maxn];
bool check(string s, string t) {
  int indx = 0;
  for (int i = 0; i < s.length(); i++) {
    if (t[indx] == s[i]) indx++;
  }
  if (indx == t.length()) return true;
  return false;
}
bool check(string s, string t1, string t2) {
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < s.length(); i++) {
    for (int j = 0; j <= t1.size(); j++) {
      if (dp[i][j] < 0) continue;
      if (j < t1.size() && s[i] == t1[j]) {
        dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      }
      if (dp[i][j] < t2.size() && s[i] == t2[dp[i][j]]) {
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + 1);
      }
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
    }
  }
  if (dp[s.length()][t1.length()] == t2.length()) return true;
  return false;
}
void solve() {
  string s, t;
  cin >> s >> t;
  if (check(s, t)) {
    cout << "YES" << endl;
    return;
  }
  for (int i = 0; i < t.length() - 1; i++) {
    string t1 = t.substr(0, i + 1);
    string t2 = t.substr(i + 1, t.size());
    if (check(s, t1, t2)) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
