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>
#pragma optimize("O2")
using namespace std;
const int N = 404;
const int inf = 8000;
int nxt[26][N];
int dp[N][N];
template <typename T>
bool check(const T &a, const T &b) {
  int n = a.size(), m = b.size();
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= m; ++j) {
      dp[i][j] = inf;
    }
  }
  dp[0][0] = -1;
  for (int i = 1; i <= n; ++i) {
    int calc = dp[i - 1][0];
    int c = a[i - 1] - 'a';
    if (calc == inf) continue;
    dp[i][0] = min(dp[i][0], nxt[c][calc + 1]);
  }
  for (int i = 1; i <= m; ++i) {
    int calc = dp[0][i - 1];
    int c = b[i - 1] - 'a';
    if (calc == inf) continue;
    dp[0][i] = nxt[c][calc + 1];
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      {
        int c = a[i - 1] - 'a';
        int calc = dp[i - 1][j];
        if (calc != inf) {
          dp[i][j] = min(dp[i][j], nxt[c][calc + 1]);
        }
      }
      {
        int c = b[j - 1] - 'a';
        int calc = dp[i][j - 1];
        if (calc != inf) {
          dp[i][j] = min(dp[i][j], nxt[c][calc + 1]);
        }
      }
    }
  }
  return dp[n][m] != inf;
}
void solve() {
  string s, t;
  cin >> s >> t;
  string_view str(t);
  int n = s.size();
  for (int c = 0; c < 26; ++c) {
    nxt[c][n] = inf;
    for (int i = n - 1; i >= 0; --i) {
      if (s[i] == char(c + 'a'))
        nxt[c][i] = i;
      else
        nxt[c][i] = nxt[c][i + 1];
    }
  };
  bool possible = false;
  int i = 0, j = 0;
  while (i < n && j < t.size()) {
    if (s[i] == t[j]) ++j;
    ++i;
  }
  if (j == t.size()) {
    cout << "YES" << '\n';
    return;
  }
  for (int i = 0; i + 1 < t.size() && !possible; ++i) {
    possible |= check(str.substr(0, i + 1), str.substr(i + 1));
  }
  cout << (possible ? "YES" : "NO") << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
