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 maxn = 5e5 + 10;
vector<long long> pref(maxn), a(maxn), b(maxn), pos[maxn];
map<long long, long long> mapa;
mt19937 eng(chrono::high_resolution_clock::now().time_since_epoch().count());
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  uniform_int_distribution<long long> dist(1, 100 * INT32_MAX);
  for (long long i = 0; i < maxn; ++i) a[i] = dist(eng);
  for (long long i = 0; i < maxn; ++i) b[i] = dist(eng);
  long long n;
  cin >> n;
  long long sol = 0;
  mapa[0] = 1;
  long long cur = 0;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    pos[x].push_back(i);
    long long cnt = pos[x].size();
    if (cnt % 3 == 1) pref[i] = pref[i - 1] ^ a[x];
    if (cnt % 3 == 2) pref[i] = pref[i - 1] ^ b[x];
    if (cnt % 3 == 0) pref[i] = pref[i - 1] ^ a[x] ^ b[x];
    if (cnt > 3) {
      long long y = pos[x][cnt - 4];
      while (cur < y - 1) {
        mapa[pref[cur]]--;
        cur++;
      }
    }
    sol += mapa[pref[i]];
    ++mapa[pref[i]];
  }
  cout << sol << '\n';
}
