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 double eps = 1e-9;
const long long inf = 1e9;
const int N = 4e2 + 5;
const int M = 1e6 + 5;
const int mod = 1e9 + 9;
int n, m, k, t, T, len, op, x, y, z;
int dp[N][N], ok, nxt[N][27];
string s1, s2;
int read() {
  int y = 0, flag = 1;
  char ch = getchar();
  while ((ch > '9' || ch < '0') && ch != '-') ch = getchar();
  if (ch == '-') flag = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') y = y * 10 + ch - '0', ch = getchar();
  return y * flag;
}
void check(string a, string b) {
  for (int i = 0; i <= len; i++)
    for (int j = 0; j <= len; j++) dp[i][j] = 401;
  dp[0][0] = 0;
  x = ((int)(a).size());
  y = ((int)(b).size());
  for (int i = 0; i <= x; i++) {
    for (int j = 0; j <= y; j++) {
      if (!i && !j) continue;
      if (i && dp[i - 1][j] < len)
        dp[i][j] = min(dp[i][j], nxt[dp[i - 1][j]][a[i - 1] - 96]);
      if (j && dp[i][j - 1] < len)
        dp[i][j] = min(dp[i][j], nxt[dp[i][j - 1]][b[j - 1] - 96]);
    }
  }
  ok = (dp[x][y] <= len);
}
void solve() {
  cin >> s1 >> s2;
  len = ((int)(s1).size());
  T = ((int)(s2).size());
  ok = 0;
  for (int i = 1; i <= 26; i++) nxt[len][i] = 401;
  for (int i = len - 1; i >= 0; i--) {
    for (int j = 1; j <= 26; j++) nxt[i][j] = nxt[i + 1][j];
    nxt[i][s1[i] - 96] = i + 1;
  }
  check(s2, "");
  for (int i = 1; i < T && !ok; i++) {
    check(s2.substr(0, i), s2.substr(i));
  }
  puts(ok ? "YES" : "NO");
}
int main() {
  int o;
  scanf("%d", &o);
  while (o--) {
    solve();
  }
  return 0;
}
