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 long long inf = 1e9 + 5;
const long long INF = 1e18 + 5;
const long long maxn = 5e5 + 5;
long long n, pr[555555], mx = -1000000001;
int sum, cnt, ans;
set<int> second;
pair<int, pair<int, int> > p[555555];
int m;
bool o[105][105];
int a[555555], l[555555], r[555555], b[555555];
int dp[405][405];
vector<int> v;
long long mod = 1;
unsigned long long bin_pow(unsigned long long a, unsigned long long n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0) {
    unsigned long long b = bin_pow(a, n / 2);
    return (b * b) % mod;
  } else {
    unsigned long long b = bin_pow(a, n / 2);
    b = (b * b) % mod;
    return (b * a) % mod;
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(NULL);
  int n;
  cin >> n;
  while (n--) {
    bool ok = 0;
    string second, t;
    cin >> second >> t;
    int m = t.size();
    int x = second.size();
    second = '!' + second;
    t = '#' + t;
    second += '!';
    t += '#';
    for (int p = 1; p <= m; p++) {
      for (int j = 0; j < 401; j++)
        for (int k = 0; k < 401; k++) dp[j][k] = -1;
      dp[0][0] = 0;
      for (int i = 0; i <= x; i++)
        for (int j = 0; j <= m; j++) {
          if (dp[i][j] < 0) {
            continue;
          }
          dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
          if (second[i + 1] == t[j + 1])
            dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
          if (t[p + dp[i][j] + 1] == second[i + 1])
            dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + 1);
        }
      if (dp[x][p] >= m - p) ok = 1;
    }
    if (ok == 0)
      cout << "NO";
    else
      cout << "YES";
    cout << endl;
  }
  return 0;
}
