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: Runtime Error)

#include <stdio.h>

int main(){
    //初期入力
    int n;
    scanf("%d",&n);
    int L[n];
    for(int i=0;i<n;i++){
        scanf("%d",&L[i]);
    }

    //選択ソート
    int cp;
    int min_num;
    for(int i=0;i<n-1;i++){
        min_num=i;
        for(int j=i+1;j<n;j++){
            if(L[j]<L[min_num]){
                min_num=j;
            }
        }
        cp=L[i];
        L[i]=L[min_num];
        L[min_num]=cp;
    }
    
    int T[n][3];
    int m=0;
    int count=0;
    for(int i=n-1;i>0;i--){
        for(int j=i-1;j>=0;j--){
            for(int f=j-1;f>=0;f--){
                if(L[j]+L[f]>L[i]){
                    T[m][0]=L[i];
                    T[m][1]=L[j];
                    T[m][2]=L[f];
                    count++;
                    m++;
                }else{
                    break;
                }
            }
        }
    }
    
    
    for(int i=0;i<m-1;i++){
        if(T[i][0]==T[i+1][0]&&T[i][1]==T[i+1][1]&&T[i][2]==T[i+1][2]){
            count--;
        }
    }
    
    printf("%d\n",count);
}