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>

int main(void)
{
  int N;
  scanf("%d", &N);
  
  int P[N];
  int R[N];
  int L[N];
  int I[N+1];
  int ans = 0;

  for(int i = 0; i < N; i++){
    int p;
    scanf("%d", &p);
    P[i] = p;
    R[i] = L[i] = I[p] =i;
  }

  for(int i = 1; i < N+1; i++){
      int idx = I[i];
      int l = idx - 1;
      int r = idx + 1;
      while(-1 < l && l != L[l])
          l = L[l];
      while(r < N && r != R[r])
          r = R[r];
      L[idx] = l;
      R[idx] = r;

      if(l != -1){
          int l2 = l-1;
          while(-1 < l2 && l2 != L[l2])
              l2 = L[l2];
          ans += i * (l-l2) * (r-idx);
      }
      if(r != N){
          int r2 = r+1;
          while(r2 < N && r2 != R[r2])
              r2 = R[r2];
          ans += i * (idx-l) * (r2-r);
      }
  }
  printf("%d", ans);
  return 0;
}