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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

fun main(args: Array<String>) {
    val N = readLine()!!.toInt()
    val P = readLine()!!.split(" ").map(String::toInt)
    println(sigoto(P, 0))
}

tailrec fun sigoto(list: List<Int>, sum: Int): Int =
    if(list.size == 2) task(list, sum) else sigoto(list.drop(1), task(list, sum))

fun task(list: List<Int>, sum: Int): Int {
    var s = 0
    for (i in 2 until list.size+1) {
        val p = list.take(i).sorted().reversed()
        s += p[1]
    }
    return sum + s
}