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;
int n, a[1 << 19], m[1 << 19], lp;
uint64_t rnd[1 << 19], ans, hsh[1 << 19];
vector<int> pv[1 << 19];
map<uint64_t, int> cnt;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  mt19937_64 rng;
  uniform_int_distribution<uint64_t> dis;
  cnt[0]++;
  for (int i = 1; i <= n; ++i) rnd[i] = dis(rng);
  for (int i = 1; i <= n; ++i) {
    pv[a[i]].push_back(i);
    hsh[i] = hsh[i - 1] - m[a[i]] * rnd[a[i]];
    m[a[i]] = (m[a[i]] + 1) % 3;
    hsh[i] += m[a[i]] * rnd[a[i]];
    if (pv[a[i]].size() > 3) {
      while (lp < pv[a[i]][pv[a[i]].size() - 4]) {
        cnt[hsh[lp]]--;
        lp++;
      }
    }
    ans += cnt[hsh[i]];
    cnt[hsh[i]]++;
  }
  cout << ans;
}
