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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    string s, t;
    cin >> s >> t;
    int n = s.size();
    int m = t.size();
    vector<vector<int> > nxt;
    nxt.resize(n);
    for (int i = 0; i < n; ++i) {
      nxt[i].assign(26, -1);
      for (int j = n - 1; j > i; --j) {
        nxt[i][s[j] - 'a'] = j;
      }
    }
    auto solve = [&](int len) {
      string s1 = t.substr(0, len);
      string s2 = t.substr(len, m - len);
      int x = s1.size();
      int y = s2.size();
      vector<vector<int> > dp;
      dp.resize(x + 2);
      for (int i = 0; i < x + 1; ++i) {
        dp[i].assign(y + 2, -1);
      }
      for (int i = 0; i < n; ++i) {
        if (s[i] == s1[0]) {
          dp[1][0] = i;
          break;
        }
      }
      for (int i = 0; i < n; ++i) {
        if (s[i] == s2[0]) {
          dp[0][1] = i;
          break;
        }
      }
      for (int i = 0; i < x + 1; ++i) {
        for (int j = 0; j < y + 1; ++j) {
          if (dp[i][j] != -1) {
            if (i < x) {
              int tmp = nxt[dp[i][j]][s1[i] - 'a'];
              if (tmp != -1 && (tmp < dp[i + 1][j] || dp[i + 1][j] == -1)) {
                dp[i + 1][j] = tmp;
              }
            }
            if (j < y) {
              int tmp = nxt[dp[i][j]][s2[j] - 'a'];
              if (tmp != -1 && (tmp < dp[i][j + 1] || dp[i][j + 1] == -1)) {
                dp[i][j + 1] = tmp;
              }
            }
          }
        }
      }
      if (dp[x][y] != -1) {
        return true;
      } else {
        return false;
      }
    };
    bool f = 0;
    for (int len = 0; len < m; ++len) {
      if (solve(len)) {
        f = 1;
        break;
      }
    }
    if (f == 1) {
      cout << "YES" << '\n';
    } else {
      cout << "NO" << '\n';
    }
  }
  return 0;
}
