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;
long long mod = 1e9 + 7;
const long long N = 100017;
string s, t;
long long n, m;
long long nxt[N][31];
bool check(string a, string b) {
  long long lenA = a.size(), lenB = b.size();
  a = '.' + a;
  b = '.' + b;
  long long dp[lenA + 175][lenB + 174];
  for (long long i = 0; i <= lenA + 111; i++)
    for (long long j = 0; j <= lenB + 111; j++) dp[i][j] = 1e9;
  dp[0][0] = 0;
  for (long long i = 0; i <= lenA; i++)
    for (long long j = 0; j <= lenB; j++) {
      if (dp[i][j] != 1e9 && i + 1 <= lenA) {
        if (nxt[dp[i][j] + 1][a[i + 1] - 'a'] != -1) {
          dp[i + 1][j] = min(dp[i + 1][j], nxt[dp[i][j] + 1][a[i + 1] - 'a']);
        }
      }
      if (dp[i][j] != 1e9 && j + 1 <= lenB) {
        if (nxt[dp[i][j] + 1][b[j + 1] - 'a'] != -1) {
          dp[i][j + 1] = min(dp[i][j + 1], nxt[dp[i][j] + 1][b[j + 1] - 'a']);
        }
      }
    }
  return (dp[lenA][lenB] != 1e9);
}
void solve() {
  cin >> s >> t;
  n = s.size();
  m = t.size();
  s = '.' + s;
  t = '.' + t;
  for (long long j = 0; j <= 26; j++)
    for (long long i = 1; i <= s.size() + 174; i++) nxt[i][j] = -1;
  for (long long j = 0; j <= 25; j++)
    for (long long i = s.size() - 1; i >= 1; i--) {
      if (s[i] - 'a' == j)
        nxt[i][j] = i;
      else
        nxt[i][j] = nxt[i + 1][j];
    }
  long long ind = 1;
  for (long long i = 1; i <= n; i++) {
    if (s[i] == t[ind]) ind++;
    if (ind == m + 1) break;
  }
  if (ind == m + 1) {
    cout << "YES\n";
    return;
  }
  string a = "";
  string b = "";
  b = t;
  b = b.erase(0, 1);
  bool ok = false;
  for (long long i = 1; i <= t.size() - 1; i++) {
    a += t[i];
    b.erase(0, 1);
    if (a.size() * b.size()) ok |= (check(a, b));
  }
  cout << (ok ? "YES" : "NO");
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t24 = 1;
  cin >> t24;
  while (t24--) {
    solve();
    cout << endl;
  }
}
