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 ll = long long;
using ld = long double;
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 5e5 + 5;
vector<int> occur[N];
int n, L;
unsigned long long r[N], H[N], ans;
map<unsigned long long, int> m;
int main() {
  srand(time(NULL));
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    r[i] = uniform_int_distribution<unsigned long long>(0, 1e18)(rng);
  m[0]++;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    occur[x].push_back(i);
    H[i] = H[i - 1] + ((int)occur[x].size() % 3 ? x[r] : -2 * x[r]);
    if ((int)occur[x].size() >= 4)
      while (L + 1 <= occur[x][(int)occur[x].size() - 4]) m[H[L++]]--;
    ans += m[H[i]]++;
  }
  printf("%llu\n", ans);
  return 0;
}
