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 ll long long
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rep1(i, n) FOR(i, 1, n+1)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; --i)
#define whole(x) (x).begin(),(x).end()
#define rwhole(x) (x).rbegin(), (x).rend()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define P pair<int, int>
#define debug(var) cerr << "[" << #var << "] " << var << endl
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define eb emplace_back
#define pr(s) cout << (s) << '\n'
const ll mod = 1000000007;
const int dx[] = {-1,0,1,0};
const int dy[] = {0,-1,0,1};


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int n;
    cin >> n;
    vl l(n);
    rep(i, n) cin >> l[i];
    int ans = 0;
    rep(i, n-2) for (int j=i+1; j<n-1; j++) for(int k=j+1; k<n; k++) {
        vector<int> v(3);
        v[0] = l[i];
        v[1] = l[j];
        v[2] = l[k];
        sort(whole(v));
        if (v[0]==v[1] || v[1]==v[2]) continue;
        if (v[2]>=v[0]+v[1]) continue;
        ans++;
    }
    
    
    pr(ans);
    return 0;
}