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 GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("-ffloat-store")
using namespace std;
clock_t time_p = clock();
void aryanc403() {}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || (a.first == b.first && a.second <= b.second));
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 1000000007L;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m;
string s, t, a, b;
bool fl;
vector<vector<long long int> > dp, bb;
long long int solve(long long int n, long long int m) {
  long long int &ans = dp[n + 1][m + 1];
  if (ans != -1) return ans;
  ans = INF;
  if (n != -1) {
    long long int cur = solve(n - 1, m) + 1;
    if (cur < INF) ans = min(ans, bb[cur][a[n] - 'a']);
  }
  if (m != -1) {
    long long int cur = solve(n, m - 1) + 1;
    if (cur < INF) ans = min(ans, bb[cur][b[m] - 'a']);
  }
  return ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> s >> t;
    s = "#" + s;
    fl = false;
    bb.clear();
    bb.resize((int)(s).size() + 5, vector<long long int>(26, INF));
    for (i = ((int)(s).size() - 1); i >= (1); --i) {
      assert(i + 1 < (int)(bb).size());
      bb[i] = bb[i + 1];
      bb[i][s[i] - 'a'] = i;
    }
    for (long long int l = 1; l <= (int)(t).size(); ++l) {
      a = t.substr(0, l);
      b = t.substr(l);
      n = (int)(a).size();
      m = (int)(b).size();
      dp.clear();
      dp.resize(n + 2, vector<long long int>(m + 2, -1));
      dp[0][0] = 0;
      if (solve(n - 1, m - 1) < (int)(s).size()) {
        fl = true;
        break;
      }
    }
    cout << (fl ? "YES" : "NO") << "\n";
  }
  aryanc403();
  return 0;
}
