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;
string s, ts;
bool F(int ml) {
  vector<vector<int>> dp(ml + 1, vector<int>(ts.size() - ml + 1, 1e9 + 7));
  vector<int> mn(ml + 1, 1e9 + 7);
  vector<int> mm(ts.size() - ml + 1, 1e9 + 7);
  dp[0][0] = -1;
  mn[0] = 0;
  mm[0] = 0;
  for (int i = 0; i < s.size(); i++) {
    vector<pair<int, int>> fomm, fomn;
    for (int j = 0; j < ml; j++) {
      if (s[i] == ts[j]) {
        if (mn[j] < 1e9 && dp[j][mn[j]] < i) {
          dp[j + 1][mn[j]] = min(i, dp[j + 1][mn[j]]);
          fomm.push_back({mn[j], j + 1});
          fomn.push_back({j + 1, mn[j]});
        }
      }
    }
    for (int j = 0; j < ts.size() - ml; j++) {
      if (s[i] == ts[j + ml]) {
        if (mm[j] < 1e9 && dp[mm[j]][j] < i) {
          dp[mm[j]][j + 1] = min(i, dp[mm[j]][j + 1]);
          fomn.push_back({mm[j], j + 1});
          fomm.push_back({j + 1, mm[j]});
        }
      }
    }
    for (int j = 0; j < fomm.size(); j++) {
      if (mm[fomm[j].first] > 1e9) {
        mm[fomm[j].first] = fomm[j].second;
      } else
        mm[fomm[j].first] = max(mm[fomm[j].first], fomm[j].second);
    }
    for (int j = 0; j < fomn.size(); j++) {
      if (mn[fomn[j].first] > 1e9)
        mn[fomn[j].first] = fomn[j].second;
      else {
        mn[fomn[j].first] = max(mn[fomn[j].first], fomn[j].second);
      }
    }
  }
  if (dp[ml][ts.size() - ml] < 1e9) return true;
  return false;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> s >> ts;
    bool f = false;
    for (int i = 0; i < ts.size(); i++) {
      if (F(i)) {
        cout << "YES\n";
        f = true;
        break;
      }
    }
    if (!f) cout << "NO\n";
  }
}
