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;
const int maxn = 1e5 + 100;
const int mo = 1000000007;
int n, T, u, v, ls, lt;
char s[440], t[440];
int dp[440][440];
bool work(int pos) {
  memset(dp, -1, sizeof dp);
  dp[0][0] = 0;
  for (int i = (1); i <= (ls); ++i) {
    for (int j = (0); j <= (pos - 1); ++j) {
      if (dp[i - 1][j] >= 0) {
        if (dp[i - 1][j] + pos <= lt && s[i] == t[pos + dp[i - 1][j]])
          dp[i][j] = dp[i - 1][j] + 1;
        else
          dp[i][j] = dp[i - 1][j];
      }
    }
    for (int j = (1); j <= (pos - 1); ++j) {
      if (s[i] == t[j]) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
    }
  }
  if (dp[ls][pos - 1] == lt - pos + 1) return true;
  return false;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    u = 0;
    scanf("%s%s", s + 1, t + 1);
    ls = strlen(s + 1), lt = strlen(t + 1);
    for (int i = (1); i <= (lt); ++i) {
      if (work(i)) {
        u = 1, printf("YES\n");
        break;
      }
    }
    if (!u) printf("NO\n");
  }
  return 0;
}
