You are given an array a consisting of n integers. We denote the subarray a[l..r] as the array [a_l, a_{l + 1}, ..., a_r] (1 ≤ l ≤ r ≤ n).

A subarray is considered good if every integer that occurs in this subarray occurs there exactly thrice. For example, the array [1, 2, 2, 2, 1, 1, 2, 2, 2] has three good subarrays:

  * a[1..6] = [1, 2, 2, 2, 1, 1]; 
  * a[2..4] = [2, 2, 2]; 
  * a[7..9] = [2, 2, 2]. 



Calculate the number of good subarrays of the given array a.

Input

The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5).

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).

Output

Print one integer — the number of good subarrays of the array a.

Examples

Input


9
1 2 2 2 1 1 2 2 2


Output


3


Input


10
1 2 3 4 1 2 3 1 2 3


Output


0


Input


12
1 2 3 4 3 4 2 1 3 4 2 1


Output


1
from collections import Counter
from random import randint
import sys
input = sys.stdin.buffer.readline

def prog():
    n = int(input())
    a = list(map(int,input().split()))
    curr_occurences = [0]*(n + 1)
    mult = [randint(1,10**20) for i in range(n + 1)]
    hash_at_idx = [0]*(n + 1)
    hashes = Counter([0])
    curr_hash = 0
    last_3 = [[] for i in range(n + 1)]
    ans = 0
    least_start = -1
    
    for i in range(n):
        x = a[i]
        if len(last_3[x]) == 3:
            new_start = last_3[x][0]
            while least_start < new_start:
                hashes[hash_at_idx[least_start]] -= 1
                least_start += 1
            del last_3[x][0]
            
        curr_hash -= curr_occurences[x]*mult[x]
        curr_occurences[x] = (curr_occurences[x] + 1) % 3
        curr_hash += curr_occurences[x]*mult[x]
        ans += hashes[curr_hash]
        hashes[curr_hash] += 1
        hash_at_idx[i] = curr_hash
        last_3[x].append(i)

    print(ans)
    
prog()
        
