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, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (y > x) x = y;
}
const long long inf = (long long)1e18 + 17;
const long long iinf = 1500000000;
const long long dx[] = {1, 0, -1, 0};
const long long dy[] = {0, 1, 0, -1};
const long long mod = 1e9 + 7;
bool test_case = 1;
void solve() {
  string a, b;
  cin >> a >> b;
  long long next[a.size() + 10][26], n1 = a.size();
  for (long long i = 0; i < 26; i++) next[n1][i] = inf;
  for (long long i = n1 - 1; i >= 0; i--) {
    for (long long j = 0; j < 26; j++) next[i][j] = next[i + 1][j];
    next[i][a[i] - 'a'] = i;
  }
  long long n = b.size();
  long long dp[n + 10][n + 10];
  dp[0][0] = 0;
  for (long long ii = 0; ii < b.size(); ii++) {
    long long l1 = ii + 1, l2 = b.size() - l1;
    for (long long i = 0; i <= l1; i++) {
      for (long long j = 0; j <= l2; j++) {
        if (i == 0 && j == 0) continue;
        dp[i][j] = inf;
        if (j != 0) {
          long long op1 = dp[i][j - 1];
          if (op1 != inf && ii + j < b.size()) {
            long long nex = b[ii + j] - 'a';
            long long pos = next[op1][nex];
            amin(dp[i][j], pos + 1);
          }
        }
        if (i != 0) {
          long long op1 = dp[i - 1][j];
          if (op1 != inf) {
            long long nex = b[i - 1] - 'a';
            long long pos = next[op1][nex];
            amin(dp[i][j], pos + 1);
          }
        }
      }
    }
    if (dp[l1][l2] != inf) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  if (test_case == true) cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
