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;
mt19937 rd(time(0));
int a[500010];
int cnt[500010];
unsigned long long has[500010];
unsigned long long val[500010];
list<int> vec[500010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  auto t = clock();
  auto now_time = [&] { return (clock() - t) / (double)CLOCKS_PER_SEC; };
  int n;
  cin >> n;
  unsigned long long anss = 1e18;
  for (int i = 1; i <= n; i++) cin >> a[i];
  {
    for (int i = 1; i <= n; i++)
      val[i] = (unsigned long long)(rd()) * (unsigned long long)(rd()),
      vec[i].clear();
    unsigned long long ans = 0;
    map<unsigned long long, int> mp;
    mp[0]++;
    int pre = 0;
    for (int i = 1; i <= n; i++) {
      int tmp = a[i];
      has[i] = has[i - 1] + val[tmp], cnt[tmp]++, vec[tmp].push_back(i);
      if (cnt[tmp] % 3 == 0) {
        has[i] -= 3 * val[tmp];
      }
      mp[has[i]]++;
      if (vec[tmp].size() > 3) {
        int now = vec[tmp].front();
        vec[tmp].pop_front();
        for (; pre < now; pre++) {
          mp[has[pre]]--;
        }
      }
      ans += mp[has[i]] - 1;
    }
    anss = min(ans, anss);
  }
  cout << anss << endl;
}
