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;
template <class T>
using vec = vector<T>;
template <class T>
using vvec = vector<vec<T>>;
template <class T>
bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
constexpr ll mod[2] = {999999937LL, 1000000007LL};
constexpr ll base = 1e9 + 9;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vec<int> A(N);
  for (auto& x : A) {
    cin >> x;
    x--;
  }
  vvec<ll> power(N, vec<ll>(2, 1));
  for (int i = 1; i < N; i++)
    for (int j = 0; j < 2; j++) power[i][j] = power[i - 1][j] * base % mod[j];
  map<vec<ll>, int> mp;
  ll ans = 0;
  mp[{0, 0}] = 1;
  vec<int> cnt(N), num(N);
  vec<ll> now = {0, 0};
  vvec<ll> hash(N + 1);
  hash[0] = now;
  int id = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < 2; j++)
      now[j] -= num[A[i]] % 3 * power[A[i]][j] % mod[j];
    cnt[A[i]]++;
    (num[A[i]] += 1) %= 3;
    for (int j = 0; j < 2; j++)
      now[j] += num[A[i]] % 3 * power[A[i]][j] % mod[j];
    while (cnt[A[i]] > 3) {
      mp[hash[id]]--;
      cnt[A[id]]--;
      id++;
    }
    if (mp.count(now)) ans += mp[now];
    hash[i + 1] = now;
    mp[now]++;
  }
  cout << ans << "\n";
}
