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: Runtime Error)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val pList = readLine()!!.split(" ").map(String::toInt)
    val dp = Array(n, {arrayOfNulls<Int>(n)})
    var count = 0
    for (i in 0 until n-1) {
        val max = Math.max(pList[i], pList[i+1])
        val second = Math.min(pList[i], pList[i+1])
        dp[i][i+1] = max
        count += second
    }

    for (x in 2..n) {
        for (i in 0 until  n-x) {
            val max = Math.max(dp[i][i+x-1]!!, dp[i+x-1][i+x]!!)
            val second = Math.min(dp[i][i+x-1]!!, dp[i+x-1][i+x]!!)
            dp[i][i+x] = max
            count += second
        }
    }
    println(count)
}