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 = 510;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
char s[N], t[N];
int nxt[N][30], f[N][N];
void solve() {
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  int n = strlen(s + 1), m = strlen(t + 1);
  for (int i = 0; i <= 25; ++i) nxt[n][i] = n + 1;
  for (int i = n - 1; i >= 0; --i) {
    for (int j = 0; j < 26; ++j) nxt[i][j] = nxt[i + 1][j];
    nxt[i][s[i + 1] - 'a'] = i + 1;
  }
  for (int k = 1; k <= m; ++k) {
    for (int i = 0; i <= m; ++i)
      for (int j = 0; j <= m; ++j) f[i][j] = n + 1;
    f[0][k - 1] = 0;
    for (int i = 0; i < k; ++i) {
      for (int j = k - 1; j <= m; ++j) {
        if (f[i][j] == n + 1) continue;
        if (i + 1 < k)
          f[i + 1][j] = min(f[i + 1][j], nxt[f[i][j]][t[i + 1] - 'a']);
        if (j + 1 <= m)
          f[i][j + 1] = min(f[i][j + 1], nxt[f[i][j]][t[j + 1] - 'a']);
      }
    }
    if (f[k - 1][m] <= n) {
      puts("YES");
      return;
    }
  }
  puts("NO");
}
int main() {
  int T = read();
  while (T--) solve();
  return 0;
}
