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;
auto random_address = [] {
  char* p = new char;
  delete p;
  return uint64_t(p);
};
const uint64_t SEED = chrono::steady_clock::now().time_since_epoch().count() *
                      (random_address() | 1);
mt19937_64 rng(SEED);
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto& x : a) cin >> x;
  vector<long long> h(n + 1, 0);
  for (int i = 0; i <= n; i++) h[i] = rng();
  long long val = 0, res = 0;
  unordered_map<long long, int> hh;
  hh[val]++;
  vector<int> freq(n + 1, 0);
  vector<queue<int>> used(n + 1);
  vector<long long> hash(n + 1, 0);
  int s = 0;
  for (int i = 0; i < n; i++) {
    int x = a[i];
    if (used[x].size() >= 3) {
      int r = used[x].front();
      used[x].pop();
      while (s < r) {
        hh[hash[s]]--;
        s++;
      }
    }
    val -= freq[x] * h[x];
    freq[x] = (freq[x] + 1) % 3;
    val += freq[x] * h[x];
    hash[i + 1] = val;
    res += hh[val]++;
    used[x].push(i + 1);
  }
  cout << res << endl;
  return 0;
}
