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;
void err() { cout << endl; }
template <class T, class... Ts>
void err(const T& arg, const Ts&... args) {
  cout << arg << ' ';
  err(args...);
}
using pII = pair<int, int>;
using ll = long long;
using db = double;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int N = 4e2 + 10;
int n, m;
char s[N], t[N];
int f[N][N];
bool gao(int mid) {
  int s1 = mid, s2 = m - mid;
  memset(f, -1, sizeof f);
  f[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = s1; j >= 0; --j) {
      f[i][j] = max(f[i][j], f[i - 1][j]);
      if (f[i][j] == s2) continue;
      if (f[i][j] != -1 && s[i] == t[mid + f[i][j] + 1]) ++f[i][j];
      if (j && s[i] == t[j]) f[i][j] = max(f[i][j], f[i - 1][j - 1]);
    }
    if (f[i][s1] == s2) return true;
  }
  return false;
}
void RUN() {
  cin >> (s + 1);
  cin >> (t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  int pos = 1;
  for (int i = 1; i <= n && pos <= m; ++i) {
    if (s[i] == t[pos]) ++pos;
  }
  if (pos > m) {
    cout << "YES"
         << "\n";
    return;
  }
  for (int i = 1; i <= m; ++i) {
    if (gao(i)) {
      cout << "YES"
           << "\n";
      return;
    }
  }
  cout << "NO"
       << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cout << fixed << setprecision(20);
  int T;
  cin >> T;
  while (T--) {
    RUN();
  }
  return 0;
}
