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;
#define REP(i, n) for(ll i=0; i<(ll)n; i++)
#define REPR(i, n) for(ll i=(ll)n-1; i>=0; i--)
#define FOR(i, m, n) for(ll i=(ll)m; i<(ll)n; i++)
#define FORR(i, m, n) for(ll i=(ll)m-1; i>=(ll)n; i--)
#define ALL(x) (x).begin(), (x).end()
#define ASC(v) sort(ALL(v));
#define DESC(v) sort(ALL(v), greater<int>());
#define UNIQ(a) sort(ALL(a));(a).resize(unique(ALL(a)) - a.begin());
#define PRINTD(x, d) cout << fixed << setprecision(d) << x << "\n";
#define DEBUG(x) cout<<#x<<": "<<x<<"\n"
const double PI = acos(-1);
ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }
bool isprime(int x){ int i; if(x<2)return 0; else if(x==2) return 1; if(x%2==0) return 0; for(i=3; i*i<=x; i+=2) if(x%i==0) return 0; return 1; }
int digsum(int n) { int r=0; while(n>0) { r+=n%10; n/=10; } return r; }
vector<string> split(string s, char p){vector<string>v;stringstream m(s);string b;while(getline(m,b,p))v.push_back(b);return v;}
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 (a>b) { a = b; return 1; } return 0; }
struct nya{ nya(){ cin.tie(0); ios::sync_with_stdio(false); } }nyan;

int main(){
    ll N; cin >> N;
    vector<ll> L(N);
    REP(n, N) cin >> L[n];

    ll ans = 0;
    REP(i, N-2) {
        FOR(j, i+1, N-1) {
            FOR(k, j+1, N) {
                if (L[i]<L[j]+L[k] && L[j]<L[i]+L[k] && L[k]<L[i]+L[j]) {
                    if (L[i]!=L[j] && L[j]!=L[k] && L[i]!=L[k]) {
                        ans++;
                    }
                }
            }
        }
    }

    cout << ans << "\n";
}