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 form[405], need[405];
int length_form, length_need;
int next_pos[405][300];
void prepare() {
  for (int i = 0; i <= length_form; i++)
    for (int let = 'a'; let <= 'z'; let++) next_pos[i][let] = 0;
  for (int i = length_form - 1; i >= 0; i--) {
    for (int let = 'a'; let <= 'z'; let++)
      next_pos[i][let] = next_pos[i + 1][let];
    next_pos[i][form[i + 1]] = i + 1;
  }
}
int no_intersect[405][405];
bool is_good(int pos) {
  int sec_length = length_need - pos;
  for (int i = 0; i <= pos; i++)
    for (int j = 0; j <= sec_length; j++) {
      if (i == 0 && j == 0) continue;
      no_intersect[i][j] = (int)1e9;
      if (i >= 1 && no_intersect[i - 1][j] != (int)1e9) {
        int nex = next_pos[no_intersect[i - 1][j]][need[i]];
        if (nex) no_intersect[i][j] = nex;
      }
      if (j >= 1 && no_intersect[i][j - 1] != (int)1e9) {
        int nex = next_pos[no_intersect[i][j - 1]][need[j + pos]];
        if (nex) no_intersect[i][j] = min(no_intersect[i][j], nex);
      }
    }
  return no_intersect[pos][sec_length] != (int)1e9;
}
void solve() {
  scanf("%s %s", form + 1, need + 1);
  length_form = strlen(form + 1);
  length_need = strlen(need + 1);
  prepare();
  for (int i = 1; i <= length_need; i++)
    if (is_good(i)) {
      printf("YES\n");
      return;
    }
  printf("NO\n");
}
int main() {
  int tester;
  scanf("%d", &tester);
  while (tester--) solve();
}
