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;
using ll = long long;
using PII = pair<int, int>;
const int mod = 998244353;
const int inf = 1 << 30;
const int maxn = 500 + 5;
int n, m, nxt[maxn][26], dp[maxn][maxn];
char s[maxn], t[maxn];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    for (int j = 0; j < 26; j++) nxt[n][j] = nxt[n + 1][j] = 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;
    }
    auto solve = [&](int x) {
      for (int i = 0; i <= x; i++) {
        for (int j = x; j <= m; j++) {
          dp[i][j] = n + 1;
        }
      }
      dp[0][x] = 0;
      for (int i = 0; i <= x; i++) {
        for (int j = x; j <= m; j++) {
          if (dp[i][j] == n + 1) continue;
          dp[i + 1][j] = min(dp[i + 1][j], nxt[dp[i][j]][t[i + 1] - 'a']);
          dp[i][j + 1] = min(dp[i][j + 1], nxt[dp[i][j]][t[j + 1] - 'a']);
        }
      }
      return dp[x][m] <= n;
    };
    bool flag = false;
    for (int i = 1; i <= m && !flag; i++) {
      flag |= solve(i);
    }
    puts(flag ? "YES" : "NO");
  }
  return 0;
}
