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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for (int i = (int)(0); i < (int)(n); ++i)
#define reps(i, n) for (int i = (int)(1); i <= (int)(n); ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; i--)
#define irep(i, m, n) for (int i = (int)(m); i < (int)(n); ++i)
#define ireps(i, m, n) for (int i = (int)(m); i <= (int)(n); ++i)
#define FOR(e, c) for (auto &e : c)
#define SORT(v, n) sort(v, v + n);
#define vsort(v) sort(v.begin(), v.end());
#define rvisort(v) sort(v.begin(), v.end(), greater<int>());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cout(d) cout<<d<<endl;
#define coutd(d) cout<<std::setprecision(10)<<d<<endl;
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
// #define int long long
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) int(x.size())

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using ul = unsigned long;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int GCD(int a, int b) { return b ? GCD(b, a%b) : a; }
int LCM(int a, int b) { return a * b / GCD(a, b); }

const int dy[] = {0, 1, 0, -1, -1, 1, 1, -1};
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};

inline bool inside(int y, int x, int H, int W) {
    return (y >= 0 && x >= 0 && y < H && x < W);
}

const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  int n; cin>>n;
  vi a(n);
  rep(i,n) {
    cin>>a[i];
    a[i]--;
  }
  vi idx(n);
  rep(i,n) idx[a[i]]=i;
  set<int> s;
  ll ans=0;
  rrep(x,n){
    int i = idx[x];
    s.insert(i);
    
    vi l(2,-1), r(2,n);
    
    auto it = s.find(i);
    rep(j,2){
      if(it == s.begin()) break;
      it--;
      l[j] = *it;
    }
    it = s.find(i);
    rep(j,2){
      it++;
      if(it == s.end()) break;
      r[j] = *it;
    }
    vi ls(2), rs(2);
    ls[0]=i-l[0]; ls[1]=l[0]-l[1];
    rs[0]=r[0]-i; rs[1]=r[1]-r[0];
    
    ans+=((ll)ls[0]*rs[1]+ls[1]*rs[0])*(x+1);
  }
  cout<<ans<<endl;
}