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;
bool isOk(string& s, string& t, int p) {
  vector<bool> vis((p + 1) * (t.size() + 1));
  int p1 = 0;
  int p2 = 0;
  int scnt = 1;
  vector<pair<int, int>> st(vis.size());
  st[0].first = 0;
  st[0].second = p;
  for (int i = 0; i < s.size(); i++) {
    for (int j = scnt - 1; j >= 0; j--) {
      bool b1 = (st[j].first < p) && (t[st[j].first] == s[i]) &&
                !vis[st[j].second * (p + 1) + st[j].first + 1];
      bool b2 = (st[j].second < t.size()) && (t[st[j].second] == s[i]) &&
                !vis[(st[j].second + 1) * (p + 1) + st[j].first];
      if (b1) vis[st[j].second * (p + 1) + st[j].first + 1] = true;
      if (b2) vis[(st[j].second + 1) * (p + 1) + st[j].first] = true;
      if (b1 && b2) {
        st[scnt] = st[j];
        st[j].first++;
        st[scnt].second++;
        scnt++;
      } else if (b1) {
        st[j].first++;
        if (st[j].first == p && st[j].second == t.size()) return true;
      } else if (b2) {
        st[j].second++;
        if (st[j].first == p && st[j].second == t.size()) return true;
      }
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  for (int q = 0; q < t; q++) {
    string s, t;
    cin >> s >> t;
    int p = t.size() - 1;
    while (p >= 0 && !isOk(s, t, p)) p--;
    cout << ((p >= 0) ? "YES" : "NO") << endl;
  }
  return 0;
}
