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 unsigned long long seed = 13331;
int n, m;
unsigned long long a[500005];
int bit[500005];
unsigned long long mypow[500005];
map<unsigned long long, int> cnt;
int st[500005], pre[500005];
int main() {
  memset(pre, -1, sizeof(pre));
  memset(st, -1, sizeof(st));
  mypow[0] = 1;
  for (int i = 1; i < 500005; i++) mypow[i] = mypow[i - 1] * seed;
  cnt[0]++;
  a[0] = 0;
  int pos = 0;
  unsigned long long hash = 0, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    int now = st[x];
    if (now != -1) now = pre[now];
    if (now != -1) now = pre[now];
    pre[i] = st[x];
    st[x] = i;
    while (pos < now) {
      cnt[a[pos]]--;
      pos++;
    }
    bit[x]++;
    bit[x] %= 3;
    if (bit[x] == 0)
      hash -= 2 * mypow[x];
    else
      hash += mypow[x];
    ans += cnt[hash];
    cnt[hash]++;
    a[i] = hash;
  }
  cout << ans << endl;
}
