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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <cstdio>
#include <vector>
#include <iostream>
#include <utility>

#define fir first
#define sec second
#define get(n) scanf("%d",&n);
#define gets(s) string s;cin >> (s);
#define All(s) (c).begin(), (c).end()
#define chmin(x, y) x = min(x, y);
#define chmax(x, y) x = max(x, y);
#define rep(i, j) for (int (i)=0;(i)<(j);(i)++)
#define repk(i, j, k) for(int (i)=(j);(i)<(k);(i)++)
#define dump(x)  cout << #x << " = " << (x) << endl;
#define debug(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
typedef long long int ll;
const ll INF = 10241024;
int N;
vector<int> P;
void input(){
    rep(i,N){
        cin >> P[i];
        cout << P[i] << endl;
    }
}

void solve(){
    input();
    int ans = 0;

    for(int l=0;l<N-1;l++){
        int fir = max(P[l], P[l+1]);
        int sec = min(P[l], P[l+1]);

        if(l == N -2 ){
            ans += sec;
            break;
        }
        for(int r=l+1;r<N;r++){
            if(P[r] <= sec)continue;
            else if(sec < P[r] && P[r] < fir){
                sec = P[r];
            }
            else if(fir < P[r]){
                sec = P[r];
                swap(sec, fir);
            }
            ans += sec;
        }
    }

    cout << ans << endl;
    return;
}
int main () {
    solve();
    return 0;
}