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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <string.h>
#include <vector>
#include <limits>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
#include <deque>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <assert.h>
#include <sstream>
#include <type_traits>
#define SET(x,y) x |= (1 << y)
#define CLEAR(x,y) x &= ~(1<< y)
#define READ(x,y) ((0u == (x & (1<<y)))?0u:1u)
#define TOGGLE(x,y) (x ^= (1<<y))
#define watch(x) cout << (#x) << " is " << (x) << endl
#define pb push_back
#define mp make_pair
#define si(n) scanf("%d",&n)
#define sf(n) scanf("%f",&n)
#define sl(n) scanf("%lld",&n)
#define slu(n) scanf("%llu",&n)
#define sd(n) scanf("%lf",&n)
#define ss(n) scanf("%s",n)
#define pnl printf("\n")
#define ll long long
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pF first
#define pS second
#define CL(a,b) memset(a,b,sizeof(a))
#define sqr(x) ((x)*(x))
#define intLIMIT numeric_limits<int>
#define longLIMIT numeric_limits<long long>
#define GCD(a,b) __gcd((a),(b))
#define dbl(x) (double)(x)
#define vi vector <int>
#define vii vector < pair <int, int> >
#define vll vector <long long>
#define allof(x) (x).begin(), (x).end()
#define fastCin cin.sync_with_stdio(0);cin.tie(0)
#define scanUnsigned(x) do{while((x=getchar())<'0'); for(x-='0'; '0'<=(_=getchar()); x=(x<<3)+(x<<1)+_-'0');}while(0)
#define printArr(a,l) cout<<#a<<": ";for(int i=0;i<(l);i++){cout<<a[i]<<" ";}cout<<endl
#define print2dArr(a,r,c) cout<<#a<<":\n";for(int i=0;i<(r);i++){for(int j=0;j<(c);j++){cout<<a[i][j]<<" ";}cout<<endl;}
char _;
using namespace std;
void PRINT (int x, int y) { for (int i = y-1; i >= 0; i--) { cout << READ(x,i); } cout << endl; }

vector<string>vec_splitter(string s){s+=',';vector<string>res;while(!s.empty()){res.push_back(s.substr(0,s.find(',')));s=s.substr(s.find(',')+1);}
return res;}
void debug_out(vector<string>__attribute__((unused))args,__attribute__((unused))int idx,__attribute__((unused))int LINE_NUM){cerr<<endl;}
template<typename Head,typename...Tail>void debug_out(vector<string>args,int idx,int LINE_NUM,Head H,Tail...T){if(idx>0)cerr<<", ";else cerr<<"Line("<<LINE_NUM<<") ";stringstream ss;ss<<H;cerr<<args[idx]<<" = "<<ss.str();debug_out(args,idx+1,LINE_NUM,T...);}
#ifdef LOCAL_DBG
#define dbg(...) debug_out(vec_splitter(#__VA_ARGS__), 0, __LINE__, __VA_ARGS__)
#else
#define dbg(...) 42
#endif

int main () {
	fastCin;
	int n;
	cin >> n;
	vi s(n);
	for (int x =0; x < n; x++) {
		cin >> s[x];
	}
	int ans = 0;
	for (int x = 0; x < n-2; x++) {
		for (int i = x+1; i < n-1; i++) {
			for (int j = i+1; j < n; j++) {
				if (s[x] == s[i] || s[x] == s[j] || s[i] == s[j]) {
					continue;
				}
				int sum = s[x]+s[i]+s[j], mx = max ({s[x], s[i], s[j]});
				if (sum-mx > mx) {
					ans++;
					// dbg(s[x], s[i], s[j]);
				}
			}
		}
	}
	cout << ans << "\n";
	return 0;
}