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 inf = 0x3f3f3f3f;
const int N = 5e5 + 100;
int last[N][5];
struct Node {
  int l, r;
  int sum, lazy;
} tree[N << 2];
void pushup(int k) {
  if (tree[k].lazy)
    tree[k].sum = 0;
  else if (tree[k].l == tree[k].r)
    tree[k].sum = 1;
  else
    tree[k].sum = tree[k << 1].sum + tree[k << 1 | 1].sum;
}
void build(int k, int l, int r) {
  tree[k].l = l;
  tree[k].r = r;
  tree[k].sum = tree[k].lazy = 0;
  if (l == r) return;
  int mid = l + r >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
}
void update(int k, int l, int r, int val) {
  if (tree[k].r < l || tree[k].l > r) return;
  if (tree[k].l >= l && tree[k].r <= r) {
    tree[k].lazy += val;
    pushup(k);
    return;
  }
  update(k << 1, l, r, val);
  update(k << 1 | 1, l, r, val);
  pushup(k);
}
int main() {
  int n;
  scanf("%d", &n);
  build(1, 1, n);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    for (int j = 4; j > 0; j--) last[x][j] = last[x][j - 1];
    last[x][0] = i;
    update(1, i, i, 0);
    update(1, last[x][1] + 1, last[x][0], 1);
    if (last[x][3] + 1 <= last[x][2]) update(1, last[x][3] + 1, last[x][2], -1);
    if (last[x][4] + 1 <= last[x][3]) update(1, last[x][4] + 1, last[x][3], 1);
    ans += tree[1].sum;
  }
  printf("%lld\n", ans);
  return 0;
}
