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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)n;++i)
#define rrep(i,n) for(int i=(int)n-1;i>=0;--i)
using namespace std;
using ll = long long;
template<typename T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T> inline bool chmin(T& a,T b){if(b<a){a=b;return 1;}return 0;}
template<typename T> vector<T> make_vec(size_t a){return vector<T>(a);}
template<typename T,typename... Ts>
auto make_vec(size_t a,Ts... ts){return vector<decltype(make_vec<T>(ts...))>(a,make_vec<T>(ts...));}
template<typename T,typename U,typename... V>
typename enable_if<is_same<T,U>::value>::type fill_v(U& u,const V... v){u=U(v...);}
template<typename T,typename U,typename... V>
typename enable_if<!is_same<T,U>::value>::type fill_v(U& u,const V... v){for(auto& e:u)fill_v<T>(e,v...);}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<int> a(n);
    rep(i, n) cin >> a[i], a[i]--;
    vector<int> idx(n);
    rep(i, n) idx[a[i]] = i;

    multiset<int> st;
    rep(i, 2) {
        st.insert(-1);
        st.insert(n);
    }
    ll ans = 0;
    rrep(i, n) {
        int now = idx[i];
        st.insert(now);
        vector<int> l(2), r(2);
        auto it = st.lower_bound(now);
        rep(j, 2) {
            ++it;
            r[j] = *it;
        }
        it = st.lower_bound(now);
        rep(j, 2) {
            --it;
            l[j] = *it;
        }
        ans += (1LL * (l[0] - l[1]) * (r[0] - now) + 1LL * (now - l[0]) * (r[1] - r[0])) * (i+1);
    }
    cout << ans << endl;
}