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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using VI = vector<int>;
using VL = vector<ll>;
#define FOR(i,a,n) for(int (i)=(a);(i)<(n);(i)++)
#define eFOR(i,a,n) for(int (i)=(a);(i)<=(n);(i)++)
#define rFOR(i,a,n) for(int (i)=(n)-1;(i)>=(a);(i)--)
#define erFOR(i,a,n) for(int (i)=(n);(i)>=(a);(i)--)
#define SORT(i) sort((i).begin(),(i).end())
#define rSORT(i,a) sort((i).begin(),(i).end(),(a))
#define all(i) (i).begin(),(i).end()
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
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; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }

int main() {
    init();

    int n;
    cin >> n;
    VL p(n), index(n);
    FOR(i, 0, n) {
        cin >> p[i];
        index[--p[i]] = i;
    }

    ll ans = 0;
    set<ll> set = { -1,n };
    rFOR(i, 0, n) {
        ll c = index[i];
        auto itr = lower_bound(all(set), c);

        ll d = *itr, e = n;
        if (d != n)e = *++itr--;
        else itr--;
        ll b = *itr, a = -1;
        if (b != -1)a = *--itr;

        ll num = (b - a) * (d - c) + (c - b) * (e - d);
        ans += num * (i + 1LL);

        set.insert(c);
    }
    cout << ans << "\n";
}