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>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
void solve();
int dbg = 1;
using namespace std;
int main() {
  int t = 1;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
int dp[405][405];
int nxt[405][26];
string second, t;
int sep;
int func(int a, int b) {
  if (dp[a][b] != -1) return dp[a][b];
  if (a == 0 && b == 0) return 0;
  dp[a][b] = second.length() + 1;
  if (a != 0) {
    int ans = func(a - 1, b);
    int c = t[a - 1] - 'a';
    if (ans < second.length())
      dp[a][b] = min(dp[a][b], nxt[ans][c] + 1);
    else
      dp[a][b] = min(dp[a][b], (int)second.length() + 1);
  }
  if (b != 0) {
    int ans = func(a, b - 1);
    int c = t[sep + b] - 'a';
    if (ans < second.length())
      dp[a][b] = min(dp[a][b], nxt[ans][c] + 1);
    else
      dp[a][b] = min(dp[a][b], (int)second.length() + 1);
  }
  return dp[a][b];
}
void solve() {
  cin >> second >> t;
  for (int i = 0; i < 26; i++)
    for (int j = second.length() - 1; j >= 0; j--)
      if (second[j] - 'a' == i)
        nxt[j][i] = j;
      else if (j == second.length() - 1)
        nxt[j][i] = second.length() + 1;
      else
        nxt[j][i] = nxt[j + 1][i];
  for (int i = 0; i < t.length(); i++) {
    memset(dp, -1, sizeof(dp));
    sep = i;
    if (func(i + 1, t.length() - i - 1) != second.length() + 1) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
