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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const ll llinf = (1ll << 61) - 1;
const double eps = 1e-6, ldeps = 1e-9;
struct Dbg {
  static constexpr auto &os = cout;
  template <class C>
  static auto dud(C *x) -> decltype(os << *x, 0);
  template <class C>
  static char dud(...);
  template <class C>
  typename enable_if<sizeof dud<C>(0) != 1, Dbg &>::type operator<<(
      const C &x) {
    os << x;
    return *this;
  }
  template <class C>
  Dbg &dump(C b, C e) {
    *this << "\n[";
    int cur = 0;
    for (C i = b; i != e and cur < 26; i++, cur++)
      *this << ", " + 2 * (i == b) << *i;
    return *this << ']';
  }
  template <class C>
  typename enable_if<sizeof dud<C>(0) == 1, Dbg &>::type operator<<(
      const C &x) {
    return dump(begin(x), end(x));
  }
  template <class C, size_t X>
  typename enable_if<!is_same<C, const char>::value, Dbg &>::type operator<<(
      C (&x)[X]) {
    return dump(begin(x), end(x));
  }
  template <class B, class C>
  Dbg &operator<<(const pair<B, C> &x) {
    return *this << '{' << x.first << ", " << x.second << '}';
  }
  template <class C, size_t Y>
  struct TP {
    void operator()(Dbg &os, const C &t) {
      TP<C, Y - 1>()(os, t);
      os << ", " << get<Y - 1>(t);
    }
  };
  template <class C>
  struct TP<C, 1> {
    void operator()(Dbg &os, const C &t) { os << get<0>(t); }
  };
  template <class... C>
  Dbg &operator<<(const tuple<C...> &t) {
    os << '{', TP<decltype(t), sizeof...(C)>()(*this, t);
    return *this << '}';
  }
  Dbg &operator<<(ostream &(*x)(std::ostream &)) {
    os << x;
    return *this;
  }
} dbg;
void err(istringstream *iss) { delete iss; }
template <class C, class... Args>
void err(istringstream *iss, const C &val, const Args &...args) {
  string name;
  *iss >> name;
  if (name.back() == ',') name.pop_back();
  dbg << name << " = " << val << "; ", err(iss, args...);
}
int T, Q, cn;
const int inf = 1000000007, mxn = 405;
char s[mxn], t[mxn];
int n, m, dp[mxn][mxn];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.precision(11);
  cin >> T;
  while (cn++ != T) {
    cin >> s >> t;
    n = int(strlen(s)), m = int(strlen(t));
    for (int i = 1; i <= m; i++) {
      for (int j = 0; j <= n; j++) {
        for (int k = 0; k <= i; k++) {
          dp[j][k] = 0;
        }
      }
      dp[0][0] = i;
      for (int j = 0; j < n; j++) {
        for (int k = 0; k <= i; k++) {
          if (dp[j][k] < i) break;
          if (s[j] == t[k]) dp[j + 1][k + 1] = max(dp[j + 1][k + 1], dp[j][k]);
          dp[j + 1][k] = max(dp[j + 1][k], (s[j] == t[dp[j][k]]) + dp[j][k]);
        }
      }
      if (dp[n][i] == m) {
        cout << "YES\n";
        goto eot;
      }
    }
    cout << "NO\n";
  eot:;
  }
}
