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 array = readLine()!!.split(" ")
	val p = Array(n){
		array[it].toInt()
	}
	
	var result = 0
	var max: Int
	var secondMax: Int
	for(l in 0 until n-1)
		for(r in l+1 until n){
			max = Math.max(p[l], p[l+1])
			secondMax = Math.min(p[l], p[l+1])
			if( 2 <= r-l )
				for(i in l+2..r){
					if( max < p[i] ){
						secondMax = max
						max = p[i]
					}else if( secondMax < p[i] )
						secondMax = p[i]
				}
			result += secondMax
		}
	
	println(result)
}