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: Wrong Answer)

#include <stdio.h>

int sankakukei(int a[], int max);
void swap3(int a, int b, int c);

int main(void){
  int i, j, k, max;
  int array[max];
  int num;

  scanf("%d", &max);
  for(int a = 0; a < max; a++){
    scanf("%d", &array[a]);
  }

  if(max < 3) num = 0;
  else num = sankakukei(array, max);

  printf("%d", num);

  return 0;
}


int sankakukei(int a[], int max){
  int i, j, k;
  int num;
  for(i = 0; i < max; i++){
    for(j = 0; j < max; j++){
      for(k = 0; k < max; k++){
        swap3(a[i], a[j], a[k]);
        if(a[i] != a[j] && a[j] != a[k] && a[k] != a[i] && a[i] < a[j] + a[k]) num++;
      }
    }
  }
  return num;
}
void swap3(int a, int b, int c){
  int n;
  if(a < b){n = a; a = b; b = n;}
  if(a < c){n = a; a = c; c = n;}
}