Problem:
Score : 200 points

Problem Statement

We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.

In how many ways can we choose three of the sticks with different lengths that can form a triangle?

That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:

L_i, L_j, and L_k are all different.

There exists a triangle whose sides have lengths L_i, L_j, and L_k.

Constraints

1 \leq N \leq 100

1 \leq L_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 \cdots L_N

Output

Print the number of ways to choose three of the sticks with different lengths that can form a triangle.

Sample Input 1

5
4 4 9 7 5

Sample Output 1

5

The following five triples (i, j, k) satisfy the conditions: (1, 3, 4), (1, 4, 5), (2, 3, 4), (2, 4, 5), and (3, 4, 5).

Sample Input 2

6
4 5 4 3 3 5

Sample Output 2

8

We have two sticks for each of the lengths 3, 4, and 5. To satisfy the first condition, we have to choose one from each length.

There is a triangle whose sides have lengths 3, 4, and 5, so we have 2 ^ 3 = 8 triples (i, j, k) that satisfy the conditions.

Sample Input 3

10
9 4 6 1 9 6 10 6 6 8

Sample Output 3

39

Sample Input 4

2
1 1

Sample Output 4

0

No triple (i, j, k) satisfies 1 \leq i < j < k \leq N, so we should print 0.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

#ifndef M_PI
#define M_PI 3.14159265358979
#endif
#define deg_to_rad(deg) (((deg) / 360) * 2 * M_PI)
#define rad_to_deg(rad) (((rad) / 2 / M_PI) * 360)

using namespace std;

typedef long long ll;
typedef vector<long long> vll;
typedef vector<int> vi;
typedef pair<long long, long long> pll;
typedef pair<int, int> pi;
typedef long double ld;
typedef pair<long double, long double> pld;

const ll INF = 1e15;
const ll MOD = 1e9 + 7;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll i, j, k;
    ll N;
    cin >> N;
    vll L(N);
    for(auto &e : L)
        cin >> e;

    ll ans = 0;
    for(i = 0; i < N; i++) {
        for(j = i + 1; j < N; j++) {
            for(k = j + 1; k < N; k++) {
                if(L[i] == L[j] || L[j] == L[k] || L[k] == L[i]) {
                    continue;
                }
                ll emax = max(L[i], max(L[j], L[k]));
                ll emin = min(L[i], min(L[j], L[k]));
                ll emid;
                if(L[i] < emax && L[i] > emin)
                    emid = L[i];
                if(L[j] < emax && L[j] > emin)
                    emid = L[j];
                if(L[k] < emax && L[k] > emin)
                    emid = L[k];
                if(emin + emid > emax) {
                    ans++;
                }
            }
        }
    }

    cout << ans << endl;

    return 0;
}