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 + 10;
struct Node {
  int l, r;
  int cover, sum;
} tr[N << 2];
int last[N][5];
void pushup(int x) {
  if (tr[x].cover) {
    tr[x].sum = 0;
  } else {
    if (tr[x].l == tr[x].r)
      tr[x].sum = 1;
    else
      tr[x].sum = tr[x << 1].sum + tr[x << 1 | 1].sum;
  }
}
void build(int i, int l, int r) {
  tr[i].l = l, tr[i].r = r;
  if (l == r) {
    tr[i].cover = 0;
    tr[i].sum = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  pushup(i);
}
void update(int i, int x, int y, int v) {
  if (x <= tr[i].l && tr[i].r <= y) {
    tr[i].cover += v;
    pushup(i);
    return;
  }
  int mid = tr[i].l + tr[i].r >> 1;
  if (x <= mid) update(i << 1, x, y, v);
  if (y > mid) update(i << 1 | 1, x, y, v);
  pushup(i);
}
int main() {
  cin.sync_with_stdio(false);
  int n;
  cin >> n;
  build(1, 1, n);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    update(1, i, i, 0);
    update(1, last[x][1] + 1, i, 1);
    if (last[x][3] + 1 <= last[x][2]) {
      update(1, last[x][3] + 1, last[x][2], -1);
    }
    if (last[x][3] >= 1) update(1, last[x][4] + 1, last[x][3], 1);
    last[x][4] = last[x][3];
    last[x][3] = last[x][2];
    last[x][2] = last[x][1];
    last[x][1] = i;
    ans += tr[1].sum;
  }
  printf("%lld\n", ans);
  return 0;
}
