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;
long long rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long cc[] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long mod = 1e9 + 7;
const long long N = 500050;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long values[N], A[N], hashes[N], cnt[N];
vector<deque<long long> > store(N);
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long test = 1;
  while (test--) {
    long long n;
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      values[i] = rng();
    }
    map<long long, long long> freq;
    long long cur_idx = 0, ans = 0;
    freq[0]++;
    for (long long i = 1; i <= n; i++) {
      cin >> A[i];
      store[A[i]].push_back(i);
      if (store[A[i]].size() > 3) {
        long long prev_idx = store[A[i]].front();
        store[A[i]].pop_front();
        while (cur_idx < prev_idx) {
          freq[hashes[cur_idx]]--;
          cur_idx++;
        }
      }
      hashes[i] =
          hashes[i - 1] + values[A[i]] * ((cnt[A[i]] + 1) % 3 - cnt[A[i]]);
      cnt[A[i]] = (cnt[A[i]] + 1) % 3;
      ans += freq[hashes[i]]++;
    }
    cout << ans << '\n';
  }
  return 0;
}
