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)

/*
author: Apoorv Singh
*/
 
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
using namespace std;
using namespace __gnu_pbds;
 
typedef vector <int> vi;
typedef pair <int, int> pii;
typedef vector <long long> vll;
typedef pair <long long, long long> pll;
#define pb push_back   
#define all(c) c.begin(), c.end()
#define For(i, a, b) for (long long i = a; i < b; ++i)
#define Forr(i, a, b) for (long long i = a; i > b; --i)
#define um unordered_map
#define F first
#define S second
#define ll long long 
#define endl "\n"
#define min(a,b) (a < b ? a : b)
#define max(a,b) (a > b ? a : b)
#define fast_io() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

#define TRACE
#ifndef ONLINE_JUDGE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
	cout << name << " : " << arg1 << endl;  
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
	const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif

template <class X>
void printarr(X arr[], int n) 
{
	for (int i = 0; i < n; ++i)
		cout << arr[i] << ' ';
	cout << endl;
}
 
template <class T, class cmp = less<T>> using ordered_set = tree<T, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;
 
ll modulo = 1e9 + 7;

ll fpow(ll a, ll n) 
{
	ll ans = 1;
	while (n) 
	{
		if (n&1) ans = (ans*a)%modulo;
		a = (a*a)%modulo; n = n >> 1;
	}
	return ans;
}

void solve() 
{
	int n; cin >> n;
	ll arr[n];
	For(i,0,n) cin >> arr[i];
	ll ans = 0;
	sort(arr, arr+n);
	For(i,0,n)
	{
		For(j,i+1,n)
		{
			if (arr[i] == arr[j]) continue;
			For(k,j+1,n)
			{
				if (arr[i] == arr[k] or arr[j] == arr[k])
					continue;
				if (arr[i] + arr[j] > arr[k]) ++ans;
			}
		}
	}				
	cout << ans << endl;
}

int main() 
{
	fast_io();
	int t = 1;
	// cin >> t;
	while (t--) solve();
	return 0;
}