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 ll long long
using namespace std;
template<class T>inline void read(T &res)
{
    char c;
    T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
    res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';
    res*=flag;
}
void gcd(ll a,ll b,ll &d,ll &x,ll &y)
{
    if(!b)
    {
        d=a;
        x=1;
        y=0;
    }
    else
    {
        gcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}
ll in(ll a,ll mod)
{
    ll d,x,y;
    gcd(a,mod,d,x,y);
    return d==1?(x+mod)%mod:-1;
}

ll qp(ll a,ll x,ll mod)
{
    ll ans=1;
    while(x)
    {
        if(x%2==1)
            ans=(ans*a)%mod;
        a=(a*a)%mod;
        x/=2;
    }
    return ans;
}
int cmp(ll a,ll b)
{
    return a<b;
}

int ok(ll a,ll b,ll c)
{
    if(a==b||a==c||b==c)
        return 0;
    if(a+b>c)
        return 1;
    return 0;
}
ll a[105];
int main()
{
    int n,ans=0;
    read(n);
    for(int i=0;i<n;i++)
        cin>>a[i];
    sort(a,a+n,cmp);
    for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++)
        {
            for(int k=j+1;k<n;k++)
                if(ok(a[i],a[j],a[k]))
                    ans++;
        }
    cout<<ans<<endl;
    return 0;

}