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 int MOD = 1e9 + 7;
const double PI = acos(-1.0);
const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
mt19937 rnd(42);
const int K = 60;
array<int, K> operator^(const array<int, K> &a, const array<int, K> &b) {
  array<int, K> c;
  for (int i = 0; i < (K); ++i) {
    c[i] = a[i] + b[i];
    c[i] %= 3;
  }
  return c;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    a[i]--;
  }
  vector<array<int, K> > rands(n);
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (K); ++j) rands[i][j] = rand() % 3;
  map<array<int, K>, int> cnt;
  vector<array<int, K> > xors(1);
  cnt[xors.back()] = 1;
  int curr = 0;
  vector<vector<int> > pos(n);
  long long int ans = 0;
  for (int i = 0; i < (n); ++i) {
    xors.push_back(xors.back() ^ rands[a[i]]);
    pos[a[i]].push_back(i);
    int len = int((pos[a[i]]).size());
    if (len >= 4) {
      while (curr <= pos[a[i]][len - 4]) {
        --cnt[xors[curr]];
        curr++;
      }
    }
    ans += cnt[xors.back()];
    cnt[xors.back()]++;
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
