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 long long inf = 1e9 + 7;
const long long INF = 1LL << 60;
const long long mod = 1e9 + 7;
const long double eps = 1e-8;
const long double pi = acos(-1.0);
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
vector<vector<long long>> calcnext(string s) {
  long long n = (long long)s.size();
  vector<vector<long long>> res(n + 1, vector<long long>(26, n));
  for (long long i = n - 1; i >= 0; --i) {
    for (long long j = 0; j < 26; j++) {
      res[i][j] = res[i + 1][j];
    }
    res[i][s[i] - 'a'] = i;
  }
  return res;
}
void solve() {
  string s, t;
  cin >> s >> t;
  long long n = (long long)s.size(), m = (long long)t.size();
  vector<vector<long long>> next = calcnext(s);
  for (long long i = 0; i < m; ++i) {
    string tb = t.substr(0, i), tf = t.substr(i);
    vector<vector<long long>> dp(n + 3, vector<long long>(n + 3, INF));
    dp[0][0] = 0;
    for (long long j = 0; j < (long long)tb.size() + 1; j++) {
      for (long long k = 0; k < (long long)tf.size() + 1; k++) {
        long long bdp = dp[j][k];
        if (bdp >= n) continue;
        if (j < (long long)tb.size() && next[bdp][tb[j] - 'a'] != n) {
          chmin(dp[j + 1][k], next[bdp][tb[j] - 'a'] + 1);
        }
        if (k < (long long)tf.size() && next[bdp][tf[k] - 'a'] != n) {
          chmin(dp[j][k + 1], next[bdp][tf[k] - 'a'] + 1);
        }
      }
    }
    if (dp[(long long)tb.size()][(long long)tf.size()] <= n) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) solve();
  return 0;
}
