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;
template <typename T>
inline long long squ(T x) {
  return (long long)x * x;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline T read() {
  T sum = 0, fg = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') fg = -1;
  for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
  return fg * sum;
}
const int maxn = 400 + 10;
char S[maxn], T[maxn];
int n, m, dp[maxn];
inline bool chk(int x) {
  dp[0] = x;
  for (int i = 1; i <= x; i++) dp[i] = -1;
  for (int i = 0; i < n; i++) {
    for (int j = x; ~j; j--) {
      if (j < x && S[i] == T[j]) chkmax(dp[j + 1], dp[j]);
      if (~dp[j] && S[i] == T[dp[j]]) ++dp[j];
    }
  }
  return dp[x] >= m;
}
inline void solve() {
  scanf("%s", S), n = strlen(S);
  scanf("%s", T), m = strlen(T);
  for (int i = 0; i < m; i++)
    if (chk(i)) {
      printf("YES\n");
      return;
    }
  printf("NO\n");
}
int main() {
  int T = read<int>();
  while (T--) solve();
  return 0;
}
