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 lens, lent, memo[405][405][405], inf = 1000000000;
char s[405], t[405];
int dp(int i, int j, int k) {
  int ret = -inf;
  if (i == lens) return ((k == lent) ? j : ret);
  if (memo[i][j][k] != -1) return memo[i][j][k];
  ret = max(ret, dp(i + 1, j, k));
  if ((j < lent) && (s[i] == t[j])) ret = max(ret, dp(i + 1, j + 1, k));
  if ((k < lent) && (s[i] == t[k])) ret = max(ret, dp(i + 1, j, k + 1));
  memo[i][j][k] = ret;
  return ret;
}
int main() {
  int tc;
  bool found;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%s %s", s, t);
    lens = strlen(s);
    lent = strlen(t);
    for (int i = 0; i <= lens; i++)
      for (int j = 0; j <= lent; j++)
        for (int k = 0; k <= lent; k++) memo[i][j][k] = -1;
    found = false;
    for (int i = 0; i < lent; i++)
      if (dp(0, 0, i) >= i) {
        puts("YES");
        found = true;
        break;
      }
    if (!found) puts("NO");
  }
  return 0;
}
