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>
#pragma GCC optimize("Ofast")
using namespace std;
const double PI = acos(-1.0);
const long long MOD = 1000000007LL;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int dp[500][500];
void solve() {
  string S, T;
  cin >> S >> T;
  int N = S.size();
  int M = T.size();
  for (int i = 0; i < (int)(M); ++i) {
    for (int j = 0; j < (int)(N + 1); ++j)
      for (int k = 0; k < (int)(M + 1); ++k) dp[j][k] = -1;
    for (int j = 0; j < (int)(N + 1); ++j) dp[j][0] = 0;
    for (int j = 0; j < (int)(N); ++j) {
      for (int k = 0; k < (int)(i + 1); ++k) {
        chmax(dp[j + 1][k], dp[j][k]);
        if (k < i && S[j] == T[k]) {
          chmax(dp[j + 1][k + 1], dp[j][k]);
        }
        if (dp[j][k] >= 0 && dp[j][k] < M - i && S[j] == T[i + dp[j][k]]) {
          chmax(dp[j + 1][k], dp[j][k] + 1);
        }
      }
    }
    if (dp[N][i] == M - i) {
      cout << "YES" << '\n';
      return;
    }
  }
  cout << "NO" << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int Q;
  cin >> Q;
  while (Q--) {
    solve();
  }
}
