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: Time Limit Exceeded)

# Parallel?

n = gets.to_i
ps = gets.split(' ').map(&:to_i)

sum = 0
memos = {}

1.upto(n - 1) do |diff|
  1.upto(n - diff) do |l|
    r = l + diff
    memo = memos["#{l}#{r - 1}"]
    pr = ps[r - 1]
    if memo
      if pr < memo[0]
        memos["#{l}#{r}"] = memo
        sum += memo[0]
      elsif memo[1] < pr
        memos["#{l}#{r}"] = [memo[1], pr]
        sum += memo[1]
      else
        memos["#{l}#{r}"] = [pr, memo[1]]
        sum += pr
      end
    else
      pl = ps[l - 1]
      if pl < pr
        memos["#{l}#{r}"] = [pl, pr]
        sum += pl
      elsif pr < pl
        memos["#{l}#{r}"] = [pr, pl]
        sum += pr
      else
        memos["#{l}#{r}"] = [pl, pr]
        sum += pl
      end
    end
  end
end

puts sum