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 int N = 0 ? 100 : 5e5 + 7;
int n;
int a[N];
unsigned long long hc[N];
unsigned long long h[N];
int cnt[N];
vector<int> arr[N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  mt19937_64 rnd(239);
  for (int x = 1; x <= n; ++x) {
    hc[x] = rnd();
  }
  map<unsigned long long, int> r;
  int start = -1;
  ++r[0];
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    h[i] = (i == 0 ? 0 : h[i - 1]) - cnt[a[i]] * hc[a[i]];
    cnt[a[i]] = (cnt[a[i]] + 1) % 3;
    h[i] += cnt[a[i]] * hc[a[i]];
    ++r[h[i]];
    arr[a[i]].push_back(i);
    if (arr[a[i]].size() > 3) {
      while (start < arr[a[i]][arr[a[i]].size() - 4]) {
        --r[start == -1 ? 0 : h[start]];
        ++start;
      }
    }
    ans += r[h[i]] - 1;
  }
  cout << ans << '\n';
  return 0;
}
