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 readList() = readLine()!!.split(" ").map{it.toInt()}
fun main(args:Array<String>){
  val N = readLine()!!.toInt()
  val P = readList()
  var answer = 0
  val maxes = hashSetOf<Int>()
  for(k in N downTo 1){
    val i = P.indexOf(k)
    var L = Array(2){-1}
    var R = Array(2){N}
    var count = 0
    for(j in i-1 downTo 0){
      if(maxes.contains(j)){
        L[1-count] = j
        count++
      }
      if(count == 2)
      {
        break
      }
    }
    count = 0 
    for(j in i+1 until N){
      if(maxes.contains(j)){
        R[count] = j
        count++
      }
      if(count == 2)
      {
        break
      }
    }
    answer += k*((L[1]-L[0])*(R[0]-i)+(i-L[1])*(R[1]-R[0]))
    maxes += i
  }
  println(answer)
}