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 i, i0, n, m, T, ans;
char s[405], t[405], a[405], b[405];
int lena, lenb, dp[405][405];
bool chk() {
  for (int i = 0; i <= lena; i++)
    for (int i0 = 0; i0 <= n; i0++) dp[i0][i] = INT_MIN / 2;
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int x = 0; x <= lena; x++) dp[i][x] = dp[i - 1][x];
    for (int x = lena - 1; x >= 0; x--) {
      if (a[x + 1] == s[i]) {
        dp[i][x + 1] = max(dp[i][x + 1], dp[i - 1][x]);
      }
    }
    for (int x = lena; x >= 0; x--) {
      if (dp[i - 1][x] + 1 <= lenb && dp[i - 1][x] + 1 >= 1 &&
          b[dp[i - 1][x] + 1] == s[i])
        dp[i][x] = max(dp[i][x], dp[i - 1][x] + 1);
    }
  }
  return dp[n][lena] == lenb;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    bool f = 0;
    n = strlen(s + 1), m = strlen(t + 1);
    for (int i = 1; i <= m; i++) {
      lena = i, lenb = m - i;
      for (int i0 = 1; i0 <= lena; i0++) a[i0] = t[i0];
      for (int i0 = 1; i0 <= lenb; i0++) b[i0] = t[lena + i0];
      if (chk()) f = 1;
    }
    if (f)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
