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 N = 50;
array<int, N> operator^(const array<int, N>& a, const array<int, N>& b) {
  array<int, N> c;
  for (int i = 0; i < int(N); i++) {
    c[i] = a[i] + b[i];
    if (c[i] >= 3) c[i] -= 3;
  }
  return c;
}
mt19937 rnd(42);
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < int(n); i++) {
    scanf("%d", &a[i]);
    --a[i];
  }
  vector<array<int, N>> nums(n);
  for (int i = 0; i < int(n); i++)
    for (int j = 0; j < int(N); j++) nums[i][j] = rnd() % 3;
  vector<vector<int>> pos(n);
  vector<array<int, N>> pr(1);
  map<array<int, N>, int> cnt;
  cnt[pr[0]] = 1;
  int cur = 0;
  long long ans = 0;
  for (int i = 0; i < int(n); i++) {
    pr.push_back(pr.back() ^ nums[a[i]]);
    pos[a[i]].push_back(i);
    if (pos[a[i]].size() >= 4) {
      while (cur <= pos[a[i]][int(pos[a[i]].size()) - 4]) {
        --cnt[pr[cur]];
        ++cur;
      }
    }
    ans += cnt[pr.back()];
    ++cnt[pr.back()];
  }
  printf("%lld\n", ans);
}
