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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};

use std::collections::BTreeSet;

fn main() {
    input! {
        n:usize,
        p:[usize;n]
    }
    let mut set = BTreeSet::new();
    let mut idx = vec![0; n + 1];
    for (i, &x) in p.iter().enumerate() {
        idx[x] = i;
    }
    let mut ans = 0_isize;
    for i in (1..=n).rev() {
        set.insert(idx[i] as isize);
        let mut left = set.range(0..idx[i] as isize);
        let mut right = set.range(idx[i] as isize + 1..);

        if let Some(t) = left.next_back() {
            let f = *left.next_back().unwrap_or(&-1);
            let k = *right.next().unwrap_or(&(n as isize));
            ans += i as isize * (*t - f) * (k - idx[i] as isize);
        }

        let mut left = set.range(0..idx[i] as isize);
        let mut right = set.range(idx[i] as isize + 1..);

        if let Some(f) = right.next() {
            let t = *right.next().unwrap_or(&(n as isize));
            let k = *left.next_back().unwrap_or(&-1);
            ans += i as isize * (t - *f) * (idx[i] as isize - k);
        }
    }
    println!("{}", ans);
}