Problem:
Score: 500 points

Problem Statement

Given is a permutation P of \{1, 2, \ldots, N\}.

For a pair (L, R) (1 \le L \lt R \le N), let X_{L, R} be the second largest value among P_L, P_{L+1}, \ldots, P_R.

Find \displaystyle \sum_{L=1}^{N-1} \sum_{R=L+1}^{N} X_{L,R}.

Constraints

2 \le N \le 10^5

1 \le P_i \le N

P_i \neq P_j (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1 P_2 \ldots P_N

Output

Print \displaystyle \sum_{L=1}^{N-1} \sum_{R=L+1}^{N} X_{L,R}.

Sample Input 1

3
2 3 1

Sample Output 1

5

X_{1, 2} = 2, X_{1, 3} = 2, and X_{2, 3} = 1, so the sum is 2 + 2 + 1 = 5.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

30

Sample Input 3

8
8 2 7 3 4 5 6 1

Sample Output 3

136

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
#define max(p,q) ((p)>(q)?(p):(q))
#define min(p,q) ((p)<(q)?(p):(q))

int main(void)
{
	int n, x, y, ans=0;
    scanf("%d", &n);
    int p[n];
    for(int i=0; i<n; i++) scanf("%d", &p[i]);

    for(int i=0; i<n-1; i++) 
    {
        x = min(p[i], p[i+1]);
        y = max(p[i], p[i+1]);
        ans += x;
        for(int j=i+2; j<n; j++)
            {
            if(x>p[j]) ans += x;
            else
            {
                if(y<p[j]) 
                {
                    ans += y;
                    x = y;
                    y = p[j];
                }
                else
                {
                ans += p[j];
                x = p[j];                  
                }
            }
        }
    }
  
  	printf("%d", ans);
 
  	return 0;
}