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,unroll-loops,-ffloat-store")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
using namespace std;
const int maxn = 5e5 + 5;
int n, pr[maxn], id[maxn];
long long a[maxn];
map<long long, int> cnt;
vector<int> pos[maxn];
long long Rand(long long l, long long h) {
  return l +
         ((long long)rand() * (RAND_MAX + 1) * (RAND_MAX + 1) * (RAND_MAX + 1) +
          (long long)rand() * (RAND_MAX + 1) * (RAND_MAX + 1) +
          (long long)rand() * (RAND_MAX + 1) + rand()) %
             (h - l + 1);
}
void readf() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    pr[i] = id[a[i]];
    id[a[i]] = i;
  }
}
void solve() {
  for (int i = 1; i <= n; ++i) pos[a[i]].emplace_back(i);
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < pos[i].size(); ++j)
      if (j < 2)
        a[pos[i][j]] = Rand(0, 1000000000000000000);
      else
        a[pos[i][j]] = a[pos[i][j - 1]] ^ a[pos[i][j - 2]];
  for (int i = 1; i <= n; ++i) a[i] ^= a[i - 1];
  int sub_pos = 0, maintain = 0;
  long long ans = 0;
  for (int i = 0; i <= n; ++i) {
    sub_pos = max(sub_pos, pr[pr[pr[i]]]);
    while (maintain < sub_pos) cnt[a[maintain++]]--;
    ans += cnt[a[i]];
    cnt[a[i]]++;
  }
  cout << ans << '\n';
}
int main() {
  srand(time(NULL));
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  readf();
  solve();
}
