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>
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; c > '9' || c < '0'; c = getchar()) {
    if (c == '-') f = -1;
  }
  for (; c >= '0' && c <= '9'; c = getchar()) {
    x = x * 10 + c - '0';
  }
  return x * f;
}
char s[405], t[405];
std::string ss, tt;
int dp[405][405], next[405][26];
inline void get_next() {
  int n = ss.size();
  for (int i = 0; i < 26; i++) next[n][i] = next[n + 1][i] = n;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++) next[i][j] = next[i + 1][j];
    next[i][s[i] - 'a'] = i;
  }
}
inline bool solve(std::string a, std::string b) {
  dp[0][0] = -1;
  for (int i = 0; i <= a.size(); i++) {
    for (int j = 0; j <= b.size(); j++) {
      if (i == 0 && j == 0) continue;
      dp[i][j] = ss.size();
      if (i)
        dp[i][j] = std::min(dp[i][j], next[dp[i - 1][j] + 1][a[i - 1] - 'a']);
      if (j)
        dp[i][j] = std::min(dp[i][j], next[dp[i][j - 1] + 1][b[j - 1] - 'a']);
    }
  }
  return dp[a.size()][b.size()] < ss.size();
}
inline bool check() {
  for (int i = 0; i < tt.size(); i++) {
    if (solve(tt.substr(0, i), tt.substr(i))) return true;
  }
  return false;
}
inline void work() {
  for (int T = read(); T; T--) {
    scanf("%s%s", s, t);
    ss = std::string(s), tt = std::string(t);
    get_next();
    if (check())
      puts("YES");
    else
      puts("NO");
  }
}
int main() {
  work();
  return 0;
}
