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>
using namespace std;
#define ALL(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<n;i++)
#define debug(n,v) cout<<#v<<":";for(int i=0;i<n;i++)cout<<v[i]<<" ";cout<<endl; 
#define INF 1000000000
#define mod 1000000007
typedef long long ll;
const ll LINF = 1001002003004005006ll;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int main(){
    int n;
    cin>>n;
    int a[n];
    vector<int> idx(n+1);
    rep(i,n){
        cin>>a[i];
        idx[a[i]]=i;
    }
    long ans=0;
    multiset<int> used;
    /*番兵
    ↓を入れないとlower_boundの結果がused.begin(),used.end()
    になることがあり、そのときbeginの一個前、endとendの後ろとか
    参照できないからこれを入れると最高に都合がよい。
    これを入れなくてもbegin,endが出てきたら場合分けでもいける。
     */
    used.insert(-1);
    used.insert(-1);
    used.insert(n);
    used.insert(n);
    for(int i=n;i>=0;i--){
        auto ite=used.lower_bound(idx[i]);
        auto nxt=ite;
        auto pre=prev(ite);
        ans+=(*pre-*prev(pre)) * (*nxt-idx[i]) * i;
        ans+=(idx[i]-*pre) * (*next(nxt)-*nxt) * i;
        used.insert(idx[i]);
    }
    cout<<ans<<endl;
    return 0;
}