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>
std::mt19937_64 mt(114514);
int32_t main() {
  int32_t n;
  scanf("%u", &n);
  std::vector<int64_t> value(n);
  for (uint32_t i = 0; i < n; i++) {
    value[i] = mt();
  }
  std::vector<int64_t> freq(n);
  std::vector<uint64_t> hash_value(n + 1);
  std::vector<std::queue<int32_t>> indices(n);
  std::unordered_map<int64_t, int32_t> count;
  count[0] = 1;
  int64_t ans = 0;
  for (uint32_t i = 0, j = 0, x; i < n; i++) {
    scanf("%u", &x);
    if (indices[--x].size() == 3) {
      uint32_t remove = indices[x].front();
      indices[x].pop();
      while (j <= remove) count[hash_value[j++]]--;
    }
    hash_value[i + 1] = hash_value[i];
    hash_value[i + 1] -= freq[x] * value[x];
    freq[x] = (freq[x] + 1) % 3;
    hash_value[i + 1] += freq[x] * value[x];
    ans += count[hash_value[i + 1]]++;
    indices[x].push(i);
  }
  printf("%lld\n", ans);
  return 0;
}
