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;
mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rofl(chrono::steady_clock::now().time_since_epoch().count());
const int N = 5e5 + 2;
const int M = 1e5;
const int mod = 0;
const int inf = 2e9 + 3;
const long long INF = 1e18;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
void files() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
int n, a[N], cnt[N];
unsigned long long g[N], pref[N], ans, hsh;
queue<int> q[N];
vector<unsigned long long> e;
int was[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    g[i] = rofl();
    cin >> a[i];
  }
  int l = 0;
  e.push_back(0);
  for (int i = 1; i <= n; ++i) {
    hsh += g[a[i]];
    cnt[a[i]]++;
    if (cnt[a[i]] == 3) {
      cnt[a[i]] = 0;
      hsh -= 3LL * g[a[i]];
    }
    pref[i] = hsh;
    e.push_back(hsh);
  }
  sort(e.begin(), e.end());
  e.erase(unique(e.begin(), e.end()), e.end());
  for (int i = 1; i <= n; ++i)
    pref[i] = lower_bound(e.begin(), e.end(), pref[i]) - e.begin();
  was[0]++;
  for (int i = 1; i <= n; ++i) {
    q[a[i]].push(i);
    if ((int)q[a[i]].size() == 4) {
      int pos = q[a[i]].front();
      q[a[i]].pop();
      while (l < pos) {
        was[pref[l]]--;
        ++l;
      }
    }
    ans += was[pref[i]];
    was[pref[i]]++;
  }
  cout << ans << '\n';
}
