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;
const int MOD = 1e9 + 7;
const int C = 26;
void add(int &x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 5e5 + 10;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
deque<int> pos[MAXN];
int n, a[MAXN], cnt[MAXN];
unsigned long long vec[MAXN], sec[MAXN];
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) vec[i] = rng();
  unordered_map<unsigned long long, int> mp;
  mp[0] = 1;
  unsigned long long ans = 0;
  unsigned long long now = 0;
  for (int i = 1, j = 0; i <= n; i++) {
    int x = a[i];
    pos[x].push_back(i);
    now -= cnt[x] * vec[x];
    cnt[x] = (cnt[x] + 1) % 3;
    now += cnt[x] * vec[x];
    sec[i] = now;
    if ((int)pos[x].size() > 3) {
      int k = pos[x].front();
      pos[x].pop_front();
      while (j < k) {
        --mp[sec[j]];
        ++j;
      }
    }
    ans += mp[now];
    mp[now]++;
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}
