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 dp[404][404];
int lastt[1000];
int nextt[404][300];
bool check(string &s, string eka, string toka) {
  int n = s.size();
  int a = eka.size();
  int b = toka.size();
  for (int i = 0; i <= a; i++) {
    for (int j = 0; j <= b; j++) {
      if (i == 0 && j == 0) continue;
      if (i == 0) {
        int ind = dp[i][j - 1];
        if (nextt[ind][toka[j - 1]] == -1)
          dp[i][j] = n + 1;
        else
          dp[i][j] = nextt[ind][toka[j - 1]] + 1;
      } else if (j == 0) {
        int ind = dp[i - 1][j];
        if (nextt[ind][eka[i - 1]] == -1)
          dp[i][j] = n + 1;
        else
          dp[i][j] = nextt[ind][eka[i - 1]] + 1;
      } else {
        dp[i][j] = n + 1;
        int ind = dp[i][j - 1];
        if (nextt[ind][toka[j - 1]] != -1)
          dp[i][j] = min(dp[i][j], nextt[ind][toka[j - 1]] + 1);
        ind = dp[i - 1][j];
        if (nextt[ind][eka[i - 1]] != -1)
          dp[i][j] = min(dp[i][j], nextt[ind][eka[i - 1]] + 1);
      }
    }
  }
  return dp[a][b] <= n;
}
void testCase() {
  string s, t;
  cin >> s >> t;
  int n = s.size();
  for (int i = 'a'; i <= 'z'; i++) lastt[i] = -1;
  for (int i = n + 1; i >= 0; i--) {
    for (int j = 'a'; j <= 'z'; j++) {
      nextt[i][j] = -1;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    lastt[s[i]] = i;
    for (int j = 'a'; j <= 'z'; j++) {
      nextt[i][j] = lastt[j];
    }
  }
  n = t.size();
  for (int i = 0; i < n; i++) {
    if (check(s, t.substr(0, i + 1), t.substr(i + 1, n - (i + 1)))) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    testCase();
  }
}
