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;
template <typename T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using MaxHeap = priority_queue<T>;
const long double PI = 3.14159265358979323846;
const long double E = 2.718281828459;
const long long MOD = 1e9 + 7;
long long M(long long n) { return (n % MOD + MOD) % MOD; }
long long pow2(long long n) { return 1ll << n; }
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int T = 1;
  cin >> T;
  for (long long(I) = (0); (I) <= ((T)-1); (I)++) {
    string s, t;
    cin >> s >> t;
    long long n = s.size(), m = t.size();
    int dp[n + 1][m + 1][m + 1];
    memset(dp, 0x7f7f7f7f, sizeof(dp));
    for (long long(i) = (0); (i) <= (m); (i)++) dp[0][0][i] = i;
    for (long long(i) = (1); (i) <= (n); (i)++) {
      for (long long(j) = (0); (j) <= (m); (j)++) {
        for (long long(k) = (0); (k) <= (m); (k)++) {
          if (j < m && s[i - 1] == t[j]) {
            dp[i][j + 1][k] = min(dp[i][j + 1][k], dp[i - 1][j][k]);
            dp[i][k][j + 1] = min(dp[i][k][j + 1], dp[i - 1][k][j]);
          }
          dp[i][j][k] = min(dp[i][j][k], dp[i - 1][j][k]);
        }
      }
    }
    bool can = false;
    for (long long(i) = (0); (i) <= (m); (i)++) {
      if (dp[n][i][m] <= i) can = true;
    }
    if (can) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}
