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 C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using static System.Console;
using static System.Math;
using System.IO;
using System.Collections.Generic;
namespace I_am_the_green_coder{
    class Inonoa{
        static void Main(){
            long n = ILongs()[0];
            long[] s = ILongs();
            long len = MathPow(2,n);

            Array.Sort(s);
            if(s[len-1]==s[len-2]){
                WriteLine("No");
                return;
            }
            for(int i=0;i<n-1;i++){
                for(long j=0;j<MathPow(2,i+1);j++){
                    if(s[len-1-j]==s[len-1-j-MathPow(2,i+1)]){
                        WriteLine("No");
                        return;
                    }
                }
            }
            WriteLine("Yes");
        }

        static long MathPow(long m, long n){
            long re = 1;
            for(int i=0;i<n;i++){
                re *= m;
            }
            return re;
        }

        static readonly long MOD = 1000000007;

        ///<summary>a_1, a_2, a_3, ...みたいなの</summary>
        static long[] ILongs(){
            string[] s = ReadLine().Split();
            long[] re = new long[s.Length];
            for(long i=0;i<s.Length;i++){
                re[i] = long.Parse(s[i]);
            }
            return re;
        }

        ///<summary>xooxoxoxxoxo...みたいなの</summary>
        static char[] IChars(){
            return ReadLine().ToCharArray();
        }

        ///<summary>n m とかをちょっとはやくかくだけ</summary>
        static string[] IStrings(){
            return ReadLine().Split();
        }

        ///<summary>n行の入力を取ったり</summary>
        static long[][] ILongLongs(long n){
            long[][] re = new long[n][];
            for(long i=0;i<n;i++){
                string[] s = ReadLine().Split();
                re[i] = new long[s.Length];
                for(long j=0;j<s.Length;j++){
                    re[i][j] = long.Parse(s[j]);
                }
            }
            return re;
        }

        ///<summary>n行の文字列を取ったり</summary>
        static char[][] ICharsChars(long n){
            char[][] re = new char[n][];
            for(long i=0;i<n;i++){
                re[i] = ReadLine().ToCharArray();
            }
            return re;
        }
    }
}