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 = 5e5 + 5;
mt19937_64 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
int n, a[N], cnt[N];
uint64_t rval[N], hsh[N];
unordered_map<uint64_t, int> fre;
queue<int> q[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), rval[i] = gen();
  int l = 0;
  int64_t ans = 0;
  fre[0]++;
  for (int r = 1; r <= n; r++) {
    q[a[r]].push(r);
    if ((int)q[a[r]].size() > 3) {
      while (l < q[a[r]].front()) fre[hsh[l]]--, l++;
      q[a[r]].pop();
    }
    hsh[r] =
        hsh[r - 1] - cnt[a[r]] * rval[a[r]] + (1 + cnt[a[r]]) % 3 * rval[a[r]];
    cnt[a[r]] = (cnt[a[r]] + 1) % 3;
    ans += fre[hsh[r]];
    fre[hsh[r]]++;
  }
  printf("%lld\n", ans);
  return 0;
}
