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 <iostream>
#include <iomanip>
#include <string>
#include <stack>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <iterator>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <regex>
using namespace std;
using pii  = pair<int,int>;
using ll=long long;
using ld=long double;
#define pb push_back
#define mp make_pair
#define sc second
#define fr first
#define stpr setprecision
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define rep(i,n) for(ll i=0;i<(n);++i)
#define Rep(i,a,b) for(ll i=(a);i<(b);++i)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define rRep(i,a,b) for(int i=a;i>=b;i--)
#define crep(i) for(char i='a';i<='z';++i)
#define psortsecond(A,N) sort(A,A+N,[](const pii &a, const pii &b){return a.second<b.second;});
#define ALL(x) (x).begin(),(x).end()
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define endl '\n'
int ctoi(const char c){
  if('0' <= c && c <= '9') return (c-'0');
  return -1;
}
ll gcd(ll a,ll b){return (b == 0 ? a : gcd(b, a%b));}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
constexpr ll MOD=1000000007;
constexpr ll INF=1000000011;
constexpr ll MOD2=998244353;
constexpr ll LINF = 1001002003004005006ll;
constexpr ld EPS=10e-8;
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; }
template<typename T> istream& operator>>(istream& is,vector<T>& v){for(auto&& x:v)is >> x;return is;}
template<typename T,typename U> istream& operator>>(istream& is, pair<T,U>& p){ is >> p.first; is >> p.second; return is;}
template<typename T,typename U> ostream& operator>>(ostream& os, const pair<T,U>& p){ os << p.first << ' ' << p.second; return os;}
template<class T> ostream& operator<<(ostream& os, vector<T>& v){
  for(auto i=begin(v); i != end(v); ++i){
    if(i !=begin(v)) os << ' ';
    os << *i;
  }
  return os;
}

set<ll> st;

int main(){
  ll N;
  pair<ll,ll> A[100007];
  cin >> N;
  rep(i,N){
    cin >> A[i].fr;
    A[i].sc=i+1;
  }
  sort(A,A+N);
  reverse(A,A+N);
  rep(i,N){
    cout << A[i].sc << " ";
  }
  cout << endl;
  st.insert(0);
  st.insert(N+1);
  ll sum = 0;
  rep(i,N){
    st.insert(A[i].sc);
    decltype(st)::iterator it = st.lower_bound(A[i].sc);
    auto a = it;auto b = it;auto c = it;auto d = it;
    b--;c++;
    if(*b==*st.begin() && *c==*st.rbegin()){
    }
    else if(*b==*st.begin()){
      d++;d++;
      sum += (*d-*c)*(*it-*b)*A[i].fr;
    }
    else if(*c==*st.rbegin()){
      a--;a--;
      sum += (*a-*b)*(*it-*c)*A[i].fr;
    }
    else{
      a--;a--;d++;d++;
      sum += (*a-*b)*(*it-*c)*A[i].fr + (*d-*c)*(*it-*b)*A[i].fr;
    }
  }
  cout << sum << endl;
}