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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

//E - Second Sum
//2019/09/09

#include <stdio.h>
#include <stdlib.h>

int whichmin(int num1,int num2) {  //小さい数字を返す関数
    if (num1<num2) {
        return num1;
    }
    else {
        return num2;
    }
}

int whichmax(int num1,int num2) {  //大きい数字を返す関数
    if (num1>num2) {
        return num1;
    }
    else {
        return num2;
    }
}

int main(void) {
    int num,temp,dummax,*permutation,i,j,ans=0;
    scanf("%d",&num);
    permutation=malloc(sizeof(int)*100001);  //メモリの動的確保
    for (i=0;i<num;i++) {
        scanf("%d",&permutation[i]);
    }
    for (i=0;i<num-1;i++) {
        for (j=i+1;j<num;j++) {
            if (j==i+1) {
                temp=whichmin(permutation[i],permutation[j]);  //2番目に大きい数
                dummax=whichmax(permutation[i],permutation[j]);  //現段階で１番大きい数
            }
            else {
                if (dummax==whichmax(dummax,permutation[j])) {  //現段階の１番大きい数が１番大きい数の時
                    temp=whichmax(temp,permutation[j]);  //２番目に大きい数はその他の数のうち大きいほう
                }
                else {  //現段階の１番の数が２番目に大きい数になったとき
                    temp=dummax;  //２番目に大きい数は現段階の１番大きい数
                    dummax=permutation[j];  //一番大きい数の更新
                }
            }
            ans+=temp;
        }
    }
    free(permutation);  //メモリの開放
    printf("%d\n",ans);
    return 0;
}