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: Time Limit Exceeded)

#include <stdio.h>

int p[100000];

int argmax(int l, int r){
  int nowmax=p[l];
  int maxindex=l;
  for(int i=l+1;i<=r;i++){
    if(p[i]>nowmax){
      maxindex=i;
      nowmax=p[i];
    }
  }
  return maxindex;
}

long secondsum(int l, int r){
  if(r-l<1) return 0;
  long ans;
  int imax,i,j;
  imax=argmax(l,r);
  int ll;
  long firstsum=0;
  int tp2nd;
  for(ll=l;ll<=imax;ll++){
    if(ll==imax) tp2nd=-1;
    else tp2nd=p[ll];
    for(j=ll+1;j<=r;j++){
      if(j!=imax && p[j]>tp2nd) tp2nd=p[j];
      if(j>=imax) firstsum+=tp2nd;
    }
  }
  if(imax==l)
    ans=secondsum(imax+1,r)+firstsum;
  else if(imax==r)
    ans=secondsum(l,imax-1)+firstsum;
  else
    ans=secondsum(l,imax-1)+secondsum(imax+1,r)+firstsum;
  return ans;
}

int main()
{
  int i,n;
  long ans;
  scanf("%d", &n);
  for(i=0;i<n;i++){
    scanf("%d", &p[i]);
  }
  ans=secondsum(0,n-1);
  printf("%ld\n",ans);  
  return 0;
}