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/7415625

def Next _N, _P, _O1, _I, o
	(_N-1).downto(0){|i|
		j = [_I[i]+1, _N].min
		j = _O1[j] while j < _N and _P[j] <= _P[i]
		o[i] = j
	}
end

def Gen _N, _P
	io = (0..._N).to_a # 0
	Next(_N, _P, io, io, io)
	o1 = io.dup # 1
	return Enumerator.new{|y|
		y << o1.dup # 1
		loop {
			Next(_N, _P, o1, io, io)
			y << io.dup # 2, 3, 4,...
		}
	}
end

N = gets.to_i
P = gets.split.map &:to_i
RU = Gen(N, P).take(2)
LU = Gen(N, P.reverse).take(2).map!{|_| _.map!{|i| N-1-i }.reverse! }

p N.times.map{|i|
	ll, l, r, rr = LU[1][i], LU[0][i], RU[0][i], RU[1][i]
	P[i] * ((i-l)*(rr-r) + (r-i)*(l-ll))
}.inject(&:+)