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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, i, j, k, h, hh, MOD = 1000000007, ans, curr;
  cin >> n;
  vector<long long> a(n + 1), ct(n + 1, 0), hash(n + 1), r(n + 1);
  for (i = 1; i <= n; i++) {
    r[i] = ((rand() % 10000) * 10000LL + (rand() % 10000)) * 10000 +
           (rand() % 10000) + 1000 + i;
  }
  h = 0;
  a[0] = 0;
  hash[0] = 0;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    h -= ct[a[i]] * r[a[i]];
    ct[a[i]] = (ct[a[i]] + 1) % 3;
    h += ct[a[i]] * r[a[i]];
    hash[i] = h;
  }
  ans = 0;
  unordered_map<long long, long long> c;
  ct.assign(n + 1, 0);
  curr = 1;
  c[0] = 1;
  for (i = 1; i <= n; i++) {
    ct[a[i]]++;
    while (ct[a[i]] > 3) {
      ct[a[curr]]--;
      c[hash[curr - 1]]--;
      curr++;
    }
    ans += c[hash[i]];
    c[hash[i]]++;
  }
  cout << ans << "\n";
  return 0;
}
