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 N = 410;
int q, f[2][N][N];
char s[N], t[N];
template <class T>
inline void read(T &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
  return;
}
int main() {
  read(q);
  while (q--) {
    scanf("%s%s", s + 1, t + 1);
    int n = strlen(s + 1), m = strlen(t + 1);
    memset(f, 0xc0, sizeof(f));
    for (register int i = 1; i <= m; ++i) f[0][0][i] = 0;
    for (register int i = 1; i <= n; ++i) {
      for (register int j = 0, w = i & 1; j <= min(i, m); ++j)
        for (register int k = j; k <= m; ++k) {
          f[w][j][k] = f[w ^ 1][j][k];
          if (f[w][j][k] >= 0 && s[i] == t[k + f[w][j][k]]) ++f[w][j][k];
          if (s[i] == t[j]) f[w][j][k] = max(f[w][j][k], f[w ^ 1][j - 1][k]);
        }
    }
    int flag = 0;
    for (register int i = 0; i <= m; ++i)
      if (f[n & 1][i][i + 1] >= m - i) flag = 1;
    printf("%s\n", flag ? "YES" : "NO");
  }
  return 0;
}
