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<algorithm>
#include<bitset>
#include<cmath>
#include<complex>
#include<cstdlib>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<string>
#include<utility>
#include<vector>

using namespace std;

#define ll long long
#define pal pair<ll,ll>
#define pad pair<double,double>
#define PB push_back
#define MP make_pair
#define F first
#define S second

const ll MOD = 1e9+7;
const ll INF = 1e18;
const double EPS = 1e-8;

long long MODPOW(long long x, long long n) {
    long long ret = 1;
    while (n > 0) {
        if (n & 1) ret = ret * x % MOD;
        x = x * x % MOD;
        n >>= 1;
    }
    return ret;
}

template<typename T,typename U>
bool chmax(T& max, const U& x){
	if(max<x){
		max = x;
		return true;
	}
	return false;
}

template<typename T,typename U>
bool chmin(T& min, const U& x){
	if(min>x){
		min = x;
		return true;
	}
	return false;
}


int main(){
	ll n,ans=0;
	cin >> n;
	vector<ll> l(n);
	for(ll i=0;i<n;i++){
		cin >> l[i];
	}

	for(ll i=0;i<n;i++){
		for(ll j=i+1;j<n;j++){
			for(ll k=j+1;k<n;k++){
				if(l[i]+l[j]>l[k] && l[k]+l[j]>l[i] && l[i]+l[k]>l[j]){
					if(l[i]!=l[j]&& l[j]!=l[k]&& l[i]!=l[k]){
						ans++;
					}
				}
			}
		}
	}
	cout << ans << endl;
}