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>
#pragma GCC optimize("-Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")
using namespace std;
using ll = long long;
using ld = long double;
const int N = 5e5 + 7;
const ll mod = 1e9 + 9;
const ll inf = 2e18;
int n;
queue<int> q[N];
int freq[N];
ll rn[N], a[N];
ll prefs[N];
map<ll, int> mp;
auto random_address = [] {
  char *p = new char;
  delete p;
  return uint64_t(p);
};
const uint64_t SEED = chrono::steady_clock::now().time_since_epoch().count() *
                      (random_address() | 1);
mt19937_64 rng(SEED);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  ll pref = 0;
  for (int i = 0; i < N; ++i) rn[i] = rng();
  int l = 1;
  mp[0] = 1ll;
  ll ans = 0;
  q[0].push(0);
  for (int i = 1; i <= n && cin >> a[i]; ++i) {
    int off = freq[a[i]], nff = ++freq[a[i]] % 3;
    freq[a[i]] = (freq[a[i]] + 1) % 3;
    ll a2 = (nff - off);
    pref = ((pref + (1ll * rn[a[i]] * a2)));
    q[a[i]].push(i);
    prefs[i] = pref;
    while (q[a[i]].size() > 3) {
      q[a[l]].pop();
      mp[prefs[l++ - 1]]--;
    }
    ans += mp[pref]++;
  }
  cout << ans;
  return 0;
}
