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;
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
const int maxn = 5e5 + 5;
int n;
long long ha[maxn], ans;
long long num[maxn], rd[maxn];
map<long long, int> mp;
queue<int> q[maxn];
signed main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) rd[i] = 1ll * rnd() * rnd() % (1ll << 60);
  int st = 0;
  mp[0]++;
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    if (q[x].size() == 3) {
      int tp = q[x].front();
      q[x].pop();
      while (st < tp) mp[ha[st++]]--;
    }
    q[x].push(i);
    ha[i] = ha[i - 1] - num[x] * rd[x];
    num[x] = (num[x] + 1) % 3;
    ha[i] += num[x] * rd[x];
    ans += mp[ha[i]], mp[ha[i]]++;
  }
  printf("%lld\n", ans);
}
