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;

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pll;
typedef vector<bool> vb;
const ll oo = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
#define sz(c) ll((c).size())
#define all(c) begin(c), end(c)
#define FOR(i,a,b) for (ll i = (a); i < (b); i++)
#define FORD(i,a,b) for (ll i = (b)-1; i >= (a); i--)
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define TR(X) ({ if(1) cerr << "TR: " << (#X) << " = " << (X) << endl; })

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	ll n; cin >> n;

	vl a(n), ai(n);
	FOR(i,0,n) cin >> a[i], a[i]--;
	FOR(i,0,n) ai[a[i]] = i;

	set<ll> indices = {-1LL,n};
	
	ll res = 0;
	FORD(x,0,n) {
		ll i = ai[x];
		auto it = indices.lower_bound(i);
		ll c = *it, b = *prev(it);
		
		ll cnt = 0;
		if (next(it) != end(indices)) {
			ll d = *next(it);
			cnt += (d-c)*(i-b);
		}
		if (prev(it) != begin(indices)) {
			ll a = *prev(prev(it));
			cnt += (b-a)*(c-i);
		}
		res += cnt*(x+1);
		indices.insert(i);
	}
	cout << res << endl;
}