1418_G. Three Occurrences
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
```python
import random

n = int(input())

a = list(map(int, input().split()))
a.insert(0, 0)

cnt = [0] * (n + 1)
cnt2 = [0] * (n + 1)
hashs = [0] * (n + 1)

r1 = [None] * (n + 1)
r2 = [None] * (n + 1)

for i in range(1, n + 1):
    r1[i] = random.randint(1, 1000000000000)
    r2[i] = random.randint(1, 1000000000000)

mpp = {
    0: 1
}

sum = 0
l = 1

ans = 0

for i in range(1, n + 1):
    if cnt[a[i]] == 1:
        sum -= r1[a[i]]
    elif cnt[a[i]] == 2:
        sum -= r2[a[i]]
        
    cnt[a[i]] += 1
    
    if cnt[a[i]] == 1:
        sum += r1[a[i]]
    elif cnt[a[i]] == 2:
        sum += r2[a[i]]
    else:
        cnt[a[i]] = 0
    
    cnt2[a[i]] += 1
    
    while cnt2[a[i]] == 4:
        mpp[hashs[l - 1]] = mpp[hashs[l - 1]] - 1
        cnt2[a[l]] -= 1
        l += 1
    
    hashs[i] = sum
    
    tans = 0
    
    if sum in mpp:
        tans += mpp[sum]
        mpp[sum] = mpp[sum] + 1
    
    else:
        mpp[sum] = 1
    
    ans += tans

print(ans)
```