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;
#define ALL(x) (x).begin(),(x).end()
#define COUT(x) cout<<(x)<<"\n"
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define REP(i, n) for(int i=0;i<n;i++)
#define YES(x) cout<<(x?"YES":"NO")<<"\n"
#define Yes(x) cout<<(x?"Yes":"No")<<"\n"
#define dump(x) cout<<#x<<" = "<<(x)<<"\n"
#define endl "\n"
using G = vector<vector<int>>;
using M = map<int,int>;
using P = pair<int,int>;
using PQ = priority_queue<int>;
using PQG = priority_queue<int,vector<int>,greater<int>>;
using V = vector<int>;
using ll = long long;
using edge = struct { int to; int cost; };
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; }
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

int main() {
  IOS;
  int n; cin >> n;
  vector<ll> l(n);
  REP(i,n) cin >> l[i];
  sort(ALL(l));

  int ans = 0;
  REP(i,n) {
    for (int j = i+1; j < n; j++) {
      for (int k = j+1; k < n; k++) {
        ll a = l[i];
        ll b = l[j];
        ll c = l[k];
        if (a == b || b == c || a == c) continue;
        if (c < a+b) ans++;
      }
    }
  }
  COUT(ans);

  return 0;
}