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 f[1005][1005], nxt[1005][105];
char s[1005], t[1005];
int q, n, m;
inline bool check(int now) {
  for (int i = 0; i <= 401; i++)
    for (int j = 0; j <= 401; j++) f[i][j] = 100000;
  f[0][now] = nxt[0][t[now] - 'a'];
  for (int i = now + 1; i <= n; i++)
    if (f[0][i - 1] < 100000) f[0][i] = nxt[f[0][i - 1] + 1][t[i] - 'a'];
  f[1][now - 1] = nxt[0][t[1] - 'a'];
  for (int i = 2; i < now; i++)
    if (f[i - 1][now - 1] < 100000)
      f[i][now - 1] = nxt[f[i - 1][now - 1] + 1][t[i] - 'a'];
  for (int i = 1; i < now; i++)
    for (int j = now; j <= n; j++) {
      if (f[i - 1][j] < 100000) {
        if (nxt[f[i - 1][j] + 1][t[i] - 'a'] <= m)
          f[i][j] = min(f[i][j], nxt[f[i - 1][j] + 1][t[i] - 'a']);
      }
      if (f[i][j - 1] < 100000) {
        if (nxt[f[i][j - 1] + 1][t[j] - 'a'] <= m)
          f[i][j] = min(f[i][j], nxt[f[i][j - 1] + 1][t[j] - 'a']);
      }
    }
  int ans = f[now - 1][n];
  if (ans <= m) return 1;
  return 0;
}
int main() {
  cin >> q;
  while (q--) {
    scanf("%s%s", s + 1, t + 1);
    m = strlen(s + 1), n = strlen(t + 1);
    for (int i = 0; i <= 401; i++)
      for (int j = 0; j < 26; j++) nxt[i][j] = 100000;
    for (int i = m; i >= 0; i--) {
      for (int j = 0; j < 26; j++) nxt[i][j] = nxt[i + 1][j];
      if (i != 0) nxt[i][s[i] - 'a'] = i;
    }
    int flag = 0;
    if (n == 1) {
      for (int i = 1; i <= m; i++)
        if (s[i] == t[n]) flag = 1;
    }
    for (int i = 2; i <= n; i++) {
      if (check(i)) flag = 1;
    }
    if (flag == 1)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
