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;
constexpr int kN = int(5E2 + 10);
string s, t;
int n, dp[kN][kN], m;
vector<int> v[26];
bool can(int pos) {
  int npos = -1, lnxt, rnxt, l = 0, r = pos;
  for (int i = 0; i <= pos; i++)
    for (int j = pos; j <= m; j++) dp[i][j] = n;
  dp[0][pos] = -1;
  for (int i = 0; i < pos; i++) {
    if (v[t[i] - 'a'].empty() || v[t[i] - 'a'].back() <= dp[i][pos])
      dp[i + 1][pos] = n;
    else
      dp[i + 1][pos] =
          *upper_bound(v[t[i] - 'a'].begin(), v[t[i] - 'a'].end(), dp[i][pos]);
  }
  for (int i = pos; i < m; i++) {
    if (v[t[i] - 'a'].empty() || v[t[i] - 'a'].back() <= dp[0][i])
      dp[0][i + 1] = n;
    else
      dp[0][i + 1] =
          *upper_bound(v[t[i] - 'a'].begin(), v[t[i] - 'a'].end(), dp[0][i]);
  }
  for (int i = 1; i <= pos; i++)
    for (int j = pos + 1; j <= m; j++) {
      if (v[t[i - 1] - 'a'].empty() || v[t[i - 1] - 'a'].back() <= dp[i - 1][j])
        lnxt = n;
      else
        lnxt = *upper_bound(v[t[i - 1] - 'a'].begin(), v[t[i - 1] - 'a'].end(),
                            dp[i - 1][j]);
      if (v[t[j - 1] - 'a'].empty() || v[t[j - 1] - 'a'].back() <= dp[i][j - 1])
        rnxt = n;
      else
        rnxt = *upper_bound(v[t[j - 1] - 'a'].begin(), v[t[j - 1] - 'a'].end(),
                            dp[i][j - 1]);
      dp[i][j] = min(lnxt, rnxt);
    }
  if (dp[pos][m] < n)
    return true;
  else
    return false;
  while (l < pos && r < n) {
    if (v[t[l] - 'a'].empty() || v[t[l] - 'a'].back() <= npos) return false;
    if (v[t[r] - 'a'].empty() || v[t[r] - 'a'].back() <= npos) return false;
    lnxt = *upper_bound(v[t[l] - 'a'].begin(), v[t[l] - 'a'].end(), npos);
    rnxt = *upper_bound(v[t[r] - 'a'].begin(), v[t[r] - 'a'].end(), npos);
    if (lnxt <= rnxt) {
      l++;
      npos = lnxt;
    } else {
      r++;
      npos = rnxt;
    }
  }
  while (l < pos) {
    if (v[t[l] - 'a'].empty() || v[t[l] - 'a'].back() <= npos) return false;
    npos = *upper_bound(v[t[l] - 'a'].begin(), v[t[l] - 'a'].end(), npos);
    l++;
  }
  while (r < n) {
    if (v[t[r] - 'a'].empty() || v[t[r] - 'a'].back() <= npos) return false;
    npos = *upper_bound(v[t[r] - 'a'].begin(), v[t[r] - 'a'].end(), npos);
    r++;
  }
  return true;
}
void solve() {
  cin >> s >> t;
  n = int(s.size());
  m = int(t.size());
  for (int i = 0; i < 26; i++) v[i].clear();
  for (int i = 0; i < n; i++) v[s[i] - 'a'].push_back(i);
  for (int i = 0; i < m; i++)
    if (can(i)) {
      cout << "YES" << endl;
      return;
    }
  cout << "NO" << endl;
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}
