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)

# Studying https://atcoder.jp/contests/abc140/submissions/7391384

N = gets.to_i
P = gets.split.map &:to_i

Pi = [nil] * N
P.each_with_index{|p, i|
	Pi[p-1] = i
}

RU = (1..N).to_a + [N, N]
LU = (-1..(N-2)).to_a + [-1, -1]

p Pi.inject(0){|sum, i|
	 l,  r = LU[i], RU[i]
	ll, rr = LU[l], RU[r]
	LU[r], RU[l] = l, r
	next sum + P[i] * ((i-l)*(rr-r) + (r-i)*(l-ll))
}