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;
constexpr int max_n = 5e5 + 5, mod = 1000696969;
int tab[max_n];
long long hash_pref[max_n], hash_pref_2[max_n], hash_pref_3[max_n],
    power[max_n], power_2[max_n], power_3[max_n];
vector<int> zliczanie[max_n];
map<long long, int> hash_cnt, hash_cnt_2, hash_cnt_3;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, lim = 0;
  cin >> n;
  power[0] = 1;
  for (int i = 1; i <= n; i++) power[i] = power[i - 1] * 500029 % mod;
  power_2[0] = 1;
  for (int i = 1; i <= n; i++) power_2[i] = power_2[i - 1] * 500041 % mod;
  power_3[0] = 1;
  for (int i = 1; i <= n; i++) power_3[i] = power_3[i - 1] * 500057 % mod;
  long long ans = 0;
  hash_cnt[0] = 1;
  hash_cnt_2[0] = 1;
  hash_cnt_3[0] = 1;
  for (int i = 1; i <= n; i++) {
    cin >> tab[i];
    zliczanie[tab[i]].push_back(i);
    while (zliczanie[tab[i]].size() > 3 &&
           lim < zliczanie[tab[i]][zliczanie[tab[i]].size() - 4]) {
      hash_cnt[hash_pref[lim]]--;
      hash_cnt_2[hash_pref_2[lim]]--;
      hash_cnt_3[hash_pref_3[lim]]--;
      lim++;
    }
    hash_pref[i] = (hash_pref[i - 1] + power[tab[i]] * 500029) % mod;
    hash_pref_2[i] = (hash_pref_2[i - 1] + power_2[tab[i]] * 500041) % mod;
    hash_pref_3[i] = (hash_pref_3[i - 1] + power_3[tab[i]] * 500057) % mod;
    if (!(zliczanie[tab[i]].size() % 3)) {
      hash_pref[i] =
          (((hash_pref[i] - power[tab[i]] * 500029 * 3) % mod) + mod) % mod;
      hash_pref_2[i] =
          (((hash_pref_2[i] - power_2[tab[i]] * 500041 * 3) % mod) + mod) % mod;
      hash_pref_3[i] =
          (((hash_pref_3[i] - power_3[tab[i]] * 500057 * 3) % mod) + mod) % mod;
    }
    ans += min({hash_cnt[hash_pref[i]], hash_cnt_2[hash_pref_2[i]],
                hash_cnt_3[hash_pref_3[i]]});
    hash_cnt[hash_pref[i]]++;
    hash_cnt_2[hash_pref_2[i]]++;
    hash_cnt_3[hash_pref_3[i]]++;
  }
  cout << ans << "\n";
  return 0;
}
