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)

/*
Dont BLUNDER
*/
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize ("-ffloat-store")
#pragma GCC optimization ("unroll-loops")
//for n2 loops
#include<bits/stdc++.h>
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;


 */
#define d long double
#define int long long
#define pii pair<int,int>
#define pb push_back
#define endl '\n'
#define pi acos(-1)

#define init(x,a) memset(x,a,sizeof(x))
#define all(c) c.begin(),c.end()
int const mod = 998244353;
int const mod1 = 1000000007;
#define l first
#define r second
int toint(const string &s) {stringstream ss; ss << s; int x; ss >> x; return x;}
string tostring ( int number ) {stringstream ss; ss << number; return ss.str();}
#define out(x) cout<<#x<<" = "<<" "<<x<<"\n";
#define trace1(x)                cerr<<#x<<": "<<x<<endl
#define trace2(x, y)             cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)          cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define haha ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
// * Using PBDS *
/*typedef tree<int, null_type, less<int>, rb_tree_tag,
            tree_order_statistics_node_update>
               os; */



int modexp(int x, int n) {
	if (n == 0) {
		return 1;
	}
	else if (n & 1) {
		return (modexp((x * x) % mod, n / 2) % mod * x % mod) % mod;
	}
	else {
		return (modexp((x * x) % mod, n / 2) % mod);
	}
}

int modinv(int n)
{
	return modexp(n, mod - 2);
}

int gcd(int a, int b) {
	if (b == 0) {
		return a;
	}
	return gcd(b, a % b);
}
bool comp(pii a, pii b) {
	return a.second < b.second;

}
struct hash_pair {

	size_t operator()(const pair<int, int>& p) const
	{

		return ((  p.second ^ p.first) << 32);
	}
};



signed main()
{
	haha
// #ifndef ONLINE_JUDGE

// 	freopen("inp.txt", "r", stdin);

// 	freopen("out1.txt", "w", stdout);
// #endif
	int n;
	cin >> n;
	int arr[n];
	for (int i = 0; i < n; i++)
	{
		cin >> arr[i];
	}
	sort(arr, arr + n);
	int c = 0;
	for (int i = 0; i < n - 2; i++) {
		for (int j = i + 1; j < n - 1; j++) {
			for (int k = j + 1; k < n; k++) {
				if (arr[i] != arr[j] && arr[j] != arr[k]) {
					if (arr[i] + arr[j] > arr[k])
						c++;
				}
			}
		}
	}
	cout << c;
















}