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;
using ii = pair<int, int>;
constexpr int MX = 5e5 + 5, M0 = 998244353, M1 = 1e9 + 7;
inline ll pw(ll base, ll exp, ll MOD) {
  ll res = 1;
  while (exp) {
    if (exp & 1) (res *= base) %= MOD;
    exp >>= 1, (base *= base) %= MOD;
  }
  return res;
}
ll h0, h1, a[MX], h[MX], cnt[MX], tot[MX];
unordered_map<ll, int> mp;
int main() {
  if (fopen("in", "r")) freopen("in", "r", stdin), freopen("out", "w", stdout);
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<int> tmp(a, a + n);
  sort(begin(tmp), end(tmp));
  tmp.resize(unique(begin(tmp), end(tmp)) - begin(tmp));
  for (int i = 0; i < n; ++i)
    a[i] = lower_bound(begin(tmp), end(tmp), a[i]) - begin(tmp);
  ll ans = 0;
  ++mp[h[0]];
  for (int i = 0, j = 0; i < n; ++i) {
    while (cnt[a[i]] == 3) --mp[h[j]], --cnt[a[j]], ++j;
    int d = (tot[a[i]] % 3 == 2 ? -2 : 1);
    (h0 += pw(3, a[i], M0) * (d + M0)) %= M0;
    (h1 += pw(5, a[i], M1) * (d + M1)) %= M1;
    h[i + 1] = h0 * M1 + h1;
    ++cnt[a[i]], ++tot[a[i]];
    ans += mp[h[i + 1]]++;
  }
  cout << ans;
}
