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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
string A, B;
string r, s, t;
vector<int> occ[300];
int memo[410][410];
int get(int i, int j) {
  if (i == -1 && j == -1) return -1;
  int &ans = memo[i + 1][j + 1];
  if (ans != -1) return ans;
  if (i == -1) {
    int b = get(i, j - 1);
    auto ib = upper_bound(occ[s[j]].begin(), occ[s[j]].end(), b);
    int fb;
    if (ib == occ[s[j]].end())
      fb = t.size();
    else
      fb = *ib;
    return ans = fb;
  } else if (j == -1) {
    int a = get(i - 1, j);
    auto ia = upper_bound(occ[r[i]].begin(), occ[r[i]].end(), a);
    int fa;
    if (ia == occ[r[i]].end())
      fa = t.size();
    else
      fa = *ia;
    return ans = fa;
  } else {
    int a = get(i - 1, j);
    auto ia = upper_bound(occ[r[i]].begin(), occ[r[i]].end(), a);
    int fa;
    if (ia == occ[r[i]].end())
      fa = t.size();
    else
      fa = *ia;
    int b = get(i, j - 1);
    auto ib = upper_bound(occ[s[j]].begin(), occ[s[j]].end(), b);
    int fb;
    if (ib == occ[s[j]].end())
      fb = t.size();
    else
      fb = *ib;
    return ans = min(fa, fb);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> t >> B;
    int n = t.size();
    for (char c = 'a'; c <= 'z'; c++) occ[c].clear();
    for (int i = 0; i < t.size(); i++) occ[t[i]].push_back(i);
    bool ans = false;
    for (int i = 0; i < B.size(); i++) {
      r = string(B.begin(), B.begin() + i);
      s = string(B.begin() + i, B.end());
      for (int x = 0; x <= r.size(); x++)
        for (int y = 0; y <= s.size(); y++) memo[x][y] = -1;
      auto it = get(r.size() - 1, s.size() - 1);
      if (it < n) {
        ans = true;
        goto leave;
      }
    }
  leave:
    if (ans)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  exit(0);
}
