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)

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,fma")
#include <algorithm>
#include <bits/stdc++.h>
#include <unordered_map>
#include <unordered_set>

using namespace std;

#define PI              acos(-1.0)
#define No              cout<<"No\n";
#define Yes             cout<<"Yes\n";
#define no              cout<<"no\n";
#define yes             cout<<"yes\n";
#define NO              cout<<"NO\n";
#define YES             cout<<"YES\n";
#define MOD             (int)998244353
#define int             long long
#define ll              long long
#define pii             pair<int, int>
#define fi               first
#define se              second
#define sf(a)           scanf("%lld", &a)
#define pf(a)           printf("%lld ", a)
#define case(a)         cout<<"Case #"<<a<<":\n";
#define FASTIO          ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define pb              emplace_back
#define mp              make_pair
#define mem(arr)        memset(arr, 0, sizeof(arr))
#define mem1(arr)       memset(arr, -1, sizeof(arr))
#define all(a)          a.begin(),a.end()
#define for0(i, n)      for (int i = 0; i < n; i++)
#define for1(i, n)      for (int i = 1; i <= n; i++)
#define loop(i,a,b)     for (int i = a; i <= b; i++)
#define loopr(i,a,b)    for (int i = a ; i>=b;i--)
#define rr              return 0
#define prec(n)         fixed<<setprecision(n)
#define maxpq           priority_queue<int>
#define minpq           priority_queue<int, vector<int>, greater<int> >
#define inf             (int)(1e18)
#define vi              vector<int>
#define endl            "\n"
#define vii             vector<pii>
#define sz(s)           s.size()
#define bits(n)         __builtin_popcount(n)
#define lcm(a,b)        (a*(b/gcd(a,b)))
#define MX              500005


int gcd(int a, int b)   {return __gcd(a,b);}

inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p & 1LL) r = modMul(r, b); b = modMul(b, b); p >>= 1LL; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }

inline bool checkBit(ll n, int i) { return n&(1LL<<i); }
inline ll setBit(ll n, int i) { return n or (1LL<<i);; }
inline ll resetBit(ll n, int i) { return n&(~(1LL<<i)); }


const double eps = 1e-9;

int dx[5] = {+1, +0, -1, -0};
int dy[5] = {+0, +1, -0, -1};
int XX[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
int YY[] = { -1, 0, 1, -1, 1, -1, 0, 1 };

bool Equal(double x, double y)
{
    return fabs(x-y)<eps; ///x==y
}

bool Greater(double x, double y)
{
    return x-eps>y; ///x>y
}

bool Lesser(double x, double y)
{
    return x+eps<y; ///x<y
}

struct point
{
    double x, y, ang;
};
///Write what you need -->
/// nt, ft, kmp, tri, sufarr, ub_lb, lis



///Template Ends Here////////////////////////////


signed main()
{
//    root = new trinode();
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);

    FASTIO

    int n;
    cin>>n;

    int a[n+1];
    for1(i, n){
        cin>>a[i];
    }

    sort(a+1, a+n+1);
    int ans = 0;
    for1(i, n){
        loop(j, i+1, n){
            loop(k, j+1, n){
                if(a[i]==a[j] || a[i]==a[k] || a[j]==a[k] || a[i]+a[j]<=a[k])
                    continue;
                ans++;
            }
        }
    }

    cout<<ans;






























//    tridel(root);
}