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>
#define rep(i, start, end) for (long long i = start; i < end; ++i)
#define repreverse(i, start, end) for (long long i = start; i >= end; --i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define len(x) ((long long)(x).size())
#define lcm(a, b) ((a) / __gcd((a), (b)) * (b))
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vllvll = vector<vll>;
using vc = vector<char>;
using pll = pair<ll, ll>;
template<class T>void print1d(T x,ll n=-1){if(n==-1)n=x.size();rep(i,0,n){cout<<x[i]<<' ';}cout<<'\n';}
template<class T>void print2d(T x,ll r=-1,ll c=-1){if(r==-1)r=x.size();if(c==-1)c=x[0].size();rep(i,0,r)print1d(x[i],c);}
template<class T, class U>bool haskey(T mp, U key) { return mp.find(key) != mp.end(); }
template<class T, class U>bool isin(T el, U container) { return find(all(container), el) != container.end(); }
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
template<class T>bool even(T n) { return ! (n & 1); }
template<class T>bool odd(T n) { return n & 1; }
template<class T>ld deg2rad(T deg) { return M_PI * deg / 180.0; }
template<class T>ld rad2deg(T rad) { return 180.0 * rad / M_PI; }
ll intpow(ll a,ll n){ll p=1;while(n){if(n&1)p*=a;a*=a;n>>=1;}return p;} 
const long double pi = M_PI;
const long long big = 1LL << 50;
const long long inf = 1LL << 60;
const long long mod = 1e9 + 7;



int main()
{
        ll N;
        cin >> N;
        vll L(N);
        rep(i, 0, N) {
                ll x;
                cin >> x;
                L[i] = x;
        }
        ll ans = 0;
        sort(all(L));
        rep(i, 0, N) {
                rep(j, i+1, N) {
                        rep(k, j+1, N) {
                                ll a = L[i];
                                ll b = L[j];
                                ll c = L[k];
                                if (a == b or b == c or c == a) continue;
                                if (a + b > c and b + c > a and c + a > b) {
                                        ++ans;
                                        //cout << i << j << k << endl;
                                }
                        }
                }
        }
        cout << ans << endl;
}