#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <typename T>
void etch(T V) {
  for (auto x : V) cout << x << ' ';
  cout << '\n';
}
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) long long int idx,
               __attribute__((unused)) long long int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, long long int idx, long long int LINE_NUM,
               Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const long long int MOD = 1e9 + 7;
const long long int MOD1 = 998244353;
const long long int N = 2e5 + 5;
const long long int INF = 1000111000111000111LL;
const long double EPS = 1e-12;
const long double PI = 3.141592653589793116;
vector<long long int> g[3001];
long long int fre[3001][3001];
long long int nc4(long long int &n) {
  long long int val = (n * (n - 1) * (n - 2) * (n - 3)) / 24;
  return val;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = (0); i < (n); i++) cin >> v[i];
  for (long long int i = (0); i < (n + 1); i++) g[i].clear();
  for (long long int i = (0); i < (n + 1); i++) {
    for (long long int j = (0); j < (n + 1); j++) fre[i][j] = 0;
  }
  for (long long int i = (0); i < (n); i++) {
    g[v[i]].push_back(i + 1);
    fre[v[i]][i + 1] = 1;
  }
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= n; j++) {
      fre[i][j] += fre[i][j - 1];
    }
  }
  long long int ans = 0;
  for (long long int x = 1; x <= n; x++) {
    long long int l = (long long int)((g[x]).size());
    ans += nc4(l);
    if ((long long int)((g[x]).size()) < 2) continue;
    for (long long int y = x + 1; y <= n; y++) {
      if ((long long int)((g[y]).size()) < 2) continue;
      for (long long int i = 0; i < l; i++) {
        for (long long int j = i + 1; j < l; j++) {
          long long int c1 = fre[y][g[x][j]] - fre[y][g[x][i]];
          long long int c2 = fre[y][n] - fre[y][g[x][j]];
          long long int c3 = fre[y][g[x][i]];
          ans += (c1 * (c2 + c3));
        }
      }
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

You are given an array a_1, a_2 ... a_n. Calculate the number of tuples (i, j, k, l) such that: 

  * 1 ≤ i < j < k < l ≤ n; 
  * a_i = a_k and a_j = a_l; 

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains a single integer n (4 ≤ n ≤ 3000) — the size of the array a.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the array a.

It's guaranteed that the sum of n in one test doesn't exceed 3000.

Output

For each test case, print the number of described tuples.

Example

Input


2
5
2 2 2 2 2
6
1 3 3 1 2 3


Output


5
2

Note

In the first test case, for any four indices i < j < k < l are valid, so the answer is the number of tuples.

In the second test case, there are 2 valid tuples: 

  * (1, 2, 4, 6): a_1 = a_4 and a_2 = a_6; 
  * (1, 3, 4, 6): a_1 = a_4 and a_3 = a_6. 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <typename T>
void etch(T V) {
  for (auto x : V) cout << x << ' ';
  cout << '\n';
}
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) long long int idx,
               __attribute__((unused)) long long int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, long long int idx, long long int LINE_NUM,
               Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const long long int MOD = 1e9 + 7;
const long long int MOD1 = 998244353;
const long long int N = 2e5 + 5;
const long long int INF = 1000111000111000111LL;
const long double EPS = 1e-12;
const long double PI = 3.141592653589793116;
vector<long long int> g[3001];
long long int fre[3001][3001];
long long int nc4(long long int &n) {
  long long int val = (n * (n - 1) * (n - 2) * (n - 3)) / 24;
  return val;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = (0); i < (n); i++) cin >> v[i];
  for (long long int i = (0); i < (n + 1); i++) g[i].clear();
  for (long long int i = (0); i < (n + 1); i++) {
    for (long long int j = (0); j < (n + 1); j++) fre[i][j] = 0;
  }
  for (long long int i = (0); i < (n); i++) {
    g[v[i]].push_back(i + 1);
    fre[v[i]][i + 1] = 1;
  }
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= n; j++) {
      fre[i][j] += fre[i][j - 1];
    }
  }
  long long int ans = 0;
  for (long long int x = 1; x <= n; x++) {
    long long int l = (long long int)((g[x]).size());
    ans += nc4(l);
    if ((long long int)((g[x]).size()) < 2) continue;
    for (long long int y = x + 1; y <= n; y++) {
      if ((long long int)((g[y]).size()) < 2) continue;
      for (long long int i = 0; i < l; i++) {
        for (long long int j = i + 1; j < l; j++) {
          long long int c1 = fre[y][g[x][j]] - fre[y][g[x][i]];
          long long int c2 = fre[y][n] - fre[y][g[x][j]];
          long long int c3 = fre[y][g[x][i]];
          ans += (c1 * (c2 + c3));
        }
      }
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```