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 maxn = 400;
const int INF = 0x3f3f3f3f;
int dp[maxn + 1][maxn + 1];
int nxt[maxn + 1][26];
string s, t;
int n, m;
bool check(int a, int b) {
  memset(dp, INF, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < (a + 1); i++)
    for (int j = 0; j < (b + 1); j++) {
      if (i && dp[i - 1][j] != INF) {
        dp[i][j] = min(dp[i][j], nxt[dp[i - 1][j]][t[i - 1] - 'a']);
      }
      if (j && dp[i][j - 1] != INF) {
        dp[i][j] = min(dp[i][j], nxt[dp[i][j - 1]][t[j + a - 1] - 'a']);
      }
    }
  return dp[a][b] != INF;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    n = ((int)s.size());
    m = ((int)t.size());
    memset(nxt, INF, sizeof(nxt));
    for (int i = (n); i >= (1); i--) {
      for (int j = 0; j < (26); j++) nxt[i - 1][j] = nxt[i][j];
      nxt[i - 1][s[i - 1] - 'a'] = i;
    }
    int ans = 0;
    for (int i = 0; i < (m); i++) {
      if (check(i, m - i)) {
        ans = 1;
        break;
      }
    }
    if (ans)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
