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
#include <bits/stdc++.h>
using namespace std;
const long long N = 1000006;
long long n, a[N], f[N], h[N], x, y, ans;
map<long long, int> m;
vector<long long> v[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    f[i] = ((long long)rand() << 32) + (rand() << 16) + rand();
  }
  m[0]++;
  for (int i = 1; i <= n; i++) {
    v[a[i]].push_back(i);
    if (v[a[i]].size() > 3) y = v[a[i]][(int)v[a[i]].size() - 4];
    while (x < y) {
      m[h[x]]--;
      x++;
    }
    h[i] = h[i - 1] + f[a[i]];
    if (v[a[i]].size() % 3 == 0) {
      h[i] -= f[a[i]] * 3;
    }
    ans += m[h[i]]++;
  }
  cout << ans;
}
