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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

'''解説動画を見て実装'''
import bisect
N = int(input())
P = [int(x)-1 for x in input().split()]

idx = [None]*N

for i, p in enumerate(P):
    idx[p] = i

ans = 0
s = []  # ordered set をリストで代用 (bisect.insort で挿入)
sidx = []  # index for set
for x in range(N-1, -1, -1):
    i = idx[x]
    c = 0  # 組み合わせの個数
    bisect.insort(s, i)
    # print(s)
    lpos = [-1]*2  # x より左側で x より大きい数の位置
    rpos = [N]*2  # x より左側で x より大きい数の位置

    # 左側の探索
    it = s.index(i)  # iterator
    for j in range(2):
        if it == 0:
            break
        it -= 1
        lpos[j] = s[it]

    # 右側の探索
    it = s.index(i)  # iterator
    for j in range(2):
        it += 1
        if it == len(s):
            break
        rpos[j] = s[it]
    ls = (i-lpos[0], lpos[0]-lpos[1])
    rs = (rpos[0]-i, rpos[1]-rpos[0])
    c = ls[0]*rs[1] + ls[1]*rs[0]

    ans += c*(x+1)
print(ans)