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;
char s[405], t[405];
char t1[405], t2[405];
int dp[405][405];
int nxt[405][26];
int main() {
  int T = 1;
  cin >> T;
  while (T--) {
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    int ls = strlen(s + 1), lt = strlen(t + 1);
    for (int i = 0; i < 26; i++) nxt[ls + 1][i] = nxt[ls + 2][i] = ls + 1;
    for (int i = ls; i >= 1; i--) {
      for (int j = 0; j < 26; j++) nxt[i][j] = nxt[i + 1][j];
      nxt[i][s[i] - 'a'] = i;
    }
    int fl = 0;
    for (int i = 1; i <= lt; i++) {
      int l1 = 0, l2 = 0;
      for (int j = 1; j <= i; j++) t1[++l1] = t[j];
      for (int j = i + 1; j <= lt; j++) t2[++l2] = t[j];
      for (int j = 0; j <= l1; j++)
        for (int k = 0; k <= l2; k++) {
          if (j == 0 && k == 0)
            dp[j][k] = 0;
          else if (j == 0)
            dp[j][k] = nxt[dp[j][k - 1] + 1][t2[k] - 'a'];
          else if (k == 0)
            dp[j][k] = nxt[dp[j - 1][k] + 1][t1[j] - 'a'];
          else
            dp[j][k] = min(nxt[dp[j][k - 1] + 1][t2[k] - 'a'],
                           nxt[dp[j - 1][k] + 1][t1[j] - 'a']);
        }
      if (dp[l1][l2] <= ls) fl = 1;
    }
    puts(fl ? "YES" : "NO");
  }
  return 0;
}
