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 Ruby (2.3.3) (judge verdict: Accepted)

N=gets.to_i
ps=[nil]+gets.split.map(&:to_i)
ps_i=[]*(N+1)
ps.each_with_index do |o,i|
    next if i==0
    ps_i[o]=i
end
cd=[0,ps_i[N],N+1]
ans=0
(1..N-1).to_a.reverse.each do |e|
    ndx=cd.bsearch_index{|x|ps_i[e]<x}
    cd.insert(ndx,ps_i[e])
    if cd[ndx+1]==N+1
        r=0
    else
        r=e*(cd[ndx+2]-cd[ndx+1])*(cd[ndx]-cd[ndx-1])
    end
    if cd[ndx-1]==0
        l=0
    else
        l=e*(cd[ndx-1]-cd[ndx-2])*(cd[ndx+1]-cd[ndx])
    end
    ans+=r+l
end
p ans