#include <bits/stdc++.h>
using namespace std;
const int SIZE = 1;
const long long INF = 9223372036854775800;
const long long inf = 2147483640;
const long double pi = 3.141595153589793;
const long long M = inf;
long long st[SIZE];
void ST(int a, int cnt) {
  st[0] = 1;
  for (int i = 1; i <= cnt; i++) {
    st[i] = (st[i - 1] * a) % M;
  }
}
long long f[SIZE];
void FCT(int cnt) {
  f[0] = 1;
  for (int i = 1; i <= cnt; i++) {
    f[i] = (f[i - 1] * i) % M;
  }
}
int z[SIZE];
void Z(string s) {
  z[0] = 0;
  int n = s.size();
  int l = 0, r = 0;
  for (int i = 1; i < n; i++) {
    z[i] = max(0, min(z[i - l], r - i));
    while (z[i] + i < n && s[z[i]] == s[z[i] + i]) z[i]++;
    if (z[i] + i > r) {
      l = i;
      r = i + z[i];
    }
  }
}
long long pow(long long a, long long b) {
  if (!b) return 1;
  if (b % 2 == 1) {
    long long res = pow(a, b - 1);
    return (a * res) % M;
  } else {
    long long res = pow(a, b / 2);
    return (res * res) % M;
  }
}
long long gsd(long long a, long long b) {
  if (!b) return a;
  return gsd(b, a % b);
}
long long a[3005];
long long cnt[3005][3005];
void solve() {
  long long n;
  cin >> n;
  for (int i = 0; i < n + 2; i++) {
    a[i] = 0;
    for (int j = 0; j < n + 2; j++) {
      cnt[i][j] = 0;
    }
  }
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]][i] = 1;
  }
  long long ans = 0;
  for (long long i = 0; i < n + 2; i++) {
    for (long long j = 1; j < n; j++) {
      cnt[i][j] += cnt[i][j - 1];
    }
  }
  for (long long i = 0; i < n; i++) {
    vector<pair<long long, long long>> c =
        vector<pair<long long, long long>>(n + 1, {0, 0});
    for (long long j = 0; j < n + 1; j++) {
      c[j].second = cnt[j][n - 1] - cnt[j][i + 1];
    }
    long long now = 0;
    for (long long j = i + 2; j < n; j++) {
      now -= c[a[j - 1]].first * c[a[j - 1]].second;
      c[a[j - 1]].first++;
      now += c[a[j - 1]].first * c[a[j - 1]].second;
      now -= c[a[j]].first * c[a[j]].second;
      c[a[j]].second--;
      now += c[a[j]].first * c[a[j]].second;
      if (a[i] == a[j]) ans += now;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

### 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>
using namespace std;
const int SIZE = 1;
const long long INF = 9223372036854775800;
const long long inf = 2147483640;
const long double pi = 3.141595153589793;
const long long M = inf;
long long st[SIZE];
void ST(int a, int cnt) {
  st[0] = 1;
  for (int i = 1; i <= cnt; i++) {
    st[i] = (st[i - 1] * a) % M;
  }
}
long long f[SIZE];
void FCT(int cnt) {
  f[0] = 1;
  for (int i = 1; i <= cnt; i++) {
    f[i] = (f[i - 1] * i) % M;
  }
}
int z[SIZE];
void Z(string s) {
  z[0] = 0;
  int n = s.size();
  int l = 0, r = 0;
  for (int i = 1; i < n; i++) {
    z[i] = max(0, min(z[i - l], r - i));
    while (z[i] + i < n && s[z[i]] == s[z[i] + i]) z[i]++;
    if (z[i] + i > r) {
      l = i;
      r = i + z[i];
    }
  }
}
long long pow(long long a, long long b) {
  if (!b) return 1;
  if (b % 2 == 1) {
    long long res = pow(a, b - 1);
    return (a * res) % M;
  } else {
    long long res = pow(a, b / 2);
    return (res * res) % M;
  }
}
long long gsd(long long a, long long b) {
  if (!b) return a;
  return gsd(b, a % b);
}
long long a[3005];
long long cnt[3005][3005];
void solve() {
  long long n;
  cin >> n;
  for (int i = 0; i < n + 2; i++) {
    a[i] = 0;
    for (int j = 0; j < n + 2; j++) {
      cnt[i][j] = 0;
    }
  }
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]][i] = 1;
  }
  long long ans = 0;
  for (long long i = 0; i < n + 2; i++) {
    for (long long j = 1; j < n; j++) {
      cnt[i][j] += cnt[i][j - 1];
    }
  }
  for (long long i = 0; i < n; i++) {
    vector<pair<long long, long long>> c =
        vector<pair<long long, long long>>(n + 1, {0, 0});
    for (long long j = 0; j < n + 1; j++) {
      c[j].second = cnt[j][n - 1] - cnt[j][i + 1];
    }
    long long now = 0;
    for (long long j = i + 2; j < n; j++) {
      now -= c[a[j - 1]].first * c[a[j - 1]].second;
      c[a[j - 1]].first++;
      now += c[a[j - 1]].first * c[a[j - 1]].second;
      now -= c[a[j]].first * c[a[j]].second;
      c[a[j]].second--;
      now += c[a[j]].first * c[a[j]].second;
      if (a[i] == a[j]) ans += now;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```