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;
string s, t;
int dp[405][405];
int nxt[405][30];
int main() {
  int tc;
  cin >> tc;
  int ns, nt;
  int idx, ii, jj;
  bool flag;
  while (tc--) {
    cin >> s >> t;
    flag = false;
    ns = s.size(), nt = t.size();
    if (nt > ns) {
      cout << "NO\n";
      continue;
    }
    for (int i = 0; i < 30; i++) nxt[ns][i] = -1;
    for (int i = ns - 1; i >= 0; i--) {
      for (int j = 0; j < 30; j++) nxt[i][j] = nxt[i + 1][j];
      nxt[i][s[i] - 'a'] = i;
    }
    for (int mid = 1; mid <= nt; mid++) {
      for (int i = 0; i <= mid; i++) {
        for (int j = mid; j <= nt; j++) dp[i][j] = 2000000005;
      }
      dp[0][mid] = 0;
      int x;
      for (int i = 0; i <= mid; i++) {
        for (int j = mid; j <= nt; j++) {
          if (dp[i][j] == 2000000005) continue;
          x = nxt[dp[i][j]][t[i] - 'a'];
          if (x != -1) dp[i + 1][j] = min(dp[i + 1][j], x + 1);
          x = nxt[dp[i][j]][t[j] - 'a'];
          if (x != -1) dp[i][j + 1] = min(dp[i][j + 1], x + 1);
        }
      }
      if (dp[mid][nt] != 2000000005) {
        flag = true;
        break;
      }
    }
    if (flag)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
