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];
int S, T, dp[405][405], nxt[405][26], pos[26];
void init() {
  for (int i = 0; i < 26; i++) pos[i] = S + 1;
  for (int i = S; i >= 0; i--) {
    for (int j = 0; j < 26; j++) nxt[i][j] = pos[j];
    if (i) pos[s[i] - 'a'] = i;
  }
}
char t1[405], t2[405];
int len1, len2;
int solve(int pos) {
  for (int i = 0; i <= T; i++)
    for (int j = 0; j <= T; j++) dp[i][j] = 0x3f3f3f3f;
  len1 = pos, len2 = T - pos;
  for (int i = 1; i <= len1; i++) t1[i] = t[i];
  for (int i = 1; i <= len2; i++) t2[i] = t[i + pos];
  dp[0][0] = 0;
  for (int i = 0; i <= len1; i++)
    for (int j = 0; j <= len2; j++) {
      if (i == 0 && j == 0) continue;
      if (i) {
        int last = dp[i - 1][j];
        if (last <= S) dp[i][j] = min(dp[i][j], nxt[last][t1[i] - 'a']);
      }
      if (j) {
        int last = dp[i][j - 1];
        if (last <= S) dp[i][j] = min(dp[i][j], nxt[last][t2[j] - 'a']);
      }
    }
  if (dp[len1][len2] <= S) return 1;
  return 0;
}
int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    scanf("%s%s", s + 1, t + 1);
    S = strlen(s + 1);
    T = strlen(t + 1);
    init();
    int flag = 0;
    for (int i = 1; i <= T; i++)
      if (solve(i)) flag = 1;
    if (flag)
      puts("YES");
    else
      puts("NO");
  }
}
