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;
using ll = long long;
const ll N = 5e5 + 5, inf = N;
int n, a[N], idx[N], pre[N], c[N];
ll ans = 0, lazy[N << 2], tree[N << 2], cnt[N << 2];
void push(int id, int l, int r) {
  if (!lazy[id]) return;
  tree[id] += lazy[id];
  if (l < r) lazy[id * 2] += lazy[id], lazy[id * 2 + 1] += lazy[id];
  lazy[id] = 0;
}
void update(int L, int R, int V, int id = 1, int l = 1, int r = n) {
  push(id, l, r);
  if (l > R || r < L || L > R) return;
  if (L <= l && r <= R)
    return lazy[id] = V, cnt[id] = (l == r) ? 1 : cnt[id], (void)push(id, l, r);
  int mid = (l + r) / 2;
  update(L, R, V, id * 2, l, mid), update(L, R, V, id * 2 + 1, mid + 1, r);
  tree[id] = max(tree[id * 2], tree[id * 2 + 1]);
  cnt[id] = (tree[id] == tree[id * 2]) * cnt[id * 2] +
            (tree[id] == tree[id * 2 + 1]) * cnt[id * 2 + 1];
}
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  update(1, n, -inf);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pre[i] = idx[a[i]];
    c[a[i]]++;
    update(i, i, inf);
    if (c[a[i]] <= 2) {
      if (c[a[i]] == 2) update(1, pre[i], inf);
      update(1, i, -inf);
    }
    if (c[a[i]] >= 3) {
      update(1, pre[pre[pre[i]]], -inf);
      update(1, pre[i], c[a[i]] == 3 ? inf : 1);
      update(1, pre[pre[i]], 1);
      update(1, i, -1);
    }
    idx[a[i]] = i;
    ans += cnt[1] * !tree[1];
  }
  cout << ans;
}
