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>
int faster_in() {
  int r = 0, c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') return -faster_in();
  for (; c > 32; r = (r << 1) + (r << 3) + c - '0', c = getchar())
    ;
  return r;
}
using namespace std;
const int INF = int(1e9 + 7);
const int tam = 410;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int testCases;
  cin >> testCases;
  while (testCases--) {
    string second, t;
    cin >> second >> t;
    int n = second.size(), m = t.size();
    int nxt[26][n + 1];
    for (int i = n; i >= 0; i--) {
      for (int c = 'a' - 'a'; c <= 'z' - 'a'; c++) {
        if (i <= n - 1) {
          nxt[c][i] = nxt[c][i + 1];
          if (second[i] - 'a' == c) nxt[c][i] = i;
        } else
          nxt[c][i] = n + 1;
      }
    }
    bool good = false;
    for (int ite = (int)0; ite < (int)m + 1; ++ite) {
      string a = t.substr(0, ite);
      string b = t.substr(ite, m - ite);
      int dp[a.size() + 1][b.size() + 1];
      for (int i = (int)0; i < (int)a.size() + 1; ++i)
        for (int j = (int)0; j < (int)b.size() + 1; ++j) dp[i][j] = INF;
      dp[0][0] = 0;
      for (int i = (int)0; i < (int)a.size() + 1; ++i) {
        for (int j = (int)0; j < (int)b.size() + 1; ++j) {
          if (dp[i][j] > n) continue;
          int len = dp[i][j];
          if (i < (int)a.size() && nxt[a[i] - 'a'][len] <= n) {
            dp[i + 1][j] = min(dp[i + 1][j], nxt[a[i] - 'a'][len] + 1);
          }
          if (j < (int)b.size() && nxt[b[j] - 'a'][len] <= n) {
            dp[i][j + 1] = min(dp[i][j + 1], nxt[b[j] - 'a'][len] + 1);
          }
        }
      }
      if (dp[a.size()][b.size()] <= n) {
        good = true;
        break;
      }
    }
    if (good) {
      cout << "YES" << '\n';
    } else
      cout << "NO" << '\n';
  }
  return 0;
}
