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 long long N = 500005;
long long n, a[N], v[N], c[N], h[N], l, ans;
map<long long, long long> mp;
vector<long long> x[N];
signed main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (!v[a[i]])
      v[a[i]] = ((long long)rand()) | ((long long)rand() << 15) |
                ((long long)rand() << 30) | ((long long)rand() << 45);
    c[i] = x[a[i]].size();
    x[a[i]].push_back(i);
    if (c[i] % 3)
      h[i] = h[i - 1] + v[a[i]];
    else
      h[i] = h[i - 1] - v[a[i]] * 2;
  }
  mp[0]++;
  for (long long i = 1; i <= n; i++) {
    if (c[i] >= 3) {
      long long p = x[a[i]][c[i] - 3];
      while (l <= p) {
        if (l) mp[h[l - 1]]--;
        l++;
      }
    }
    ans += mp[h[i]];
    mp[h[i]]++;
  }
  cout << ans << endl;
  return 0;
}
