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)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define pb push_back
#define f(i,n) for(int i=0; i<n; ++i)
#define fi(i,st,n) for(int i=st; i<=n; ++i)
#define veci vector<int>
#define vecp vector<pair<int,int> >
#define vecl vector<ll>
int prime[100000+10];

ll lcm(ll a, ll b) {
	return a*b/__gcd(a,b);
}

ll power(ll a, ll n, ll mod) {
	ll res = 1;
	while(n > 0) {
		if(n&1) {
			res = (res*a)%mod;
		}
		a = (a*a)%mod;
		n = n/2;
	}
	return res;
}

ll sum(int arr[], int n) {
	ll res = 0;
	f(i,n) {
		res += arr[i];
	}
	return res;
}

void seive() {
	prime[1] = 0;
	for(int i=2; i<=100000; i++) {
		prime[i] = 1;
	}
	for(ll i=2; i<=100000; i++) {
		if(prime[i]) {
			for(ll j=i*i; j<=100000; j+=i) {
				prime[j] = 0;
			}
		}
	}
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int t = 1;
//	cin>>t;
	while(t--){
		ll n;
		cin>>n;
		int arr[n];
		f(i,n) cin>>arr[i];
		sort(arr,arr+n);
		ll ans = 0;
		for(int i=0; i<n; i++) {
			for(int j=i+1; j<n; j++) {
				for(int k=j+1; k<n; k++) {
					if(arr[k] < arr[i] + arr[j] and arr[i] != arr[j] and arr[j] != arr[k]) {
						ans++;
					}
				}
			}
		}
		cout<<ans<<"\n";
 	}
 	return 0;
}