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)

class BIT
    def initialize(n)
        @n = n
        @dat = Array.new(@n, 0)
    end

    def sum(i)
        s = 0
        while i >= 0 do
            s += @dat[i]
            i = (i & (i+1)) - 1
        end
        s
    end

    def sum_between(a, b)
        sum(b) - sum(a-1)
    end

    def add(i, x)
        while i < @n do
            @dat[i] += x
            i |= i+1
        end
    end

    def lower_bound(x)
        return -1 if x <= 0
        ret = -1
        k = 1
        k <<= 1 while 2*k <= @n
        while k > 0
            if ret+k < @n && @dat[ret+k] < x
                x -= @dat[ret+k]
                ret += k
            end
            k >>= 1
        end
        ret + 1
    end
end

N = gets.to_i
P = gets.split.map(&:to_i)
pos = []
N.times do |i|
    pos[P[i]] = i
end
bit = BIT.new(N)
ans = 0
N.downto(1) do |i|
    bit.add(pos[i], 1)
    s = bit.sum(pos[i])
    l2 = bit.lower_bound(s-2)
    l1 = bit.lower_bound(s-1)
    r1 = bit.lower_bound(s+1)
    r2 = bit.lower_bound(s+2)
    ans += i * ((l1-l2)*(r1-pos[i]) + (pos[i]-l1)*(r2-r1))
end
puts ans