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;
struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
struct ListNode {
  int val;
  ListNode* next;
  ListNode(int x) : val(x), next(NULL) {}
};
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long nxtLong() {
  long long x;
  cin >> x;
  return x;
}
string nxtStr() {
  string x;
  cin >> x;
  return x;
}
char nxtChar() {
  char x;
  cin >> x;
  return x;
}
double nxtDouble() {
  double x;
  cin >> x;
  return x;
}
long double nxtLDouble() {
  long double x;
  cin >> x;
  return x;
}
TreeNode* makeTree(int n) {
  queue<TreeNode*> tree_nodes;
  TreeNode* root;
  if (n) {
    root = new TreeNode(nxt());
    tree_nodes.push(root);
  } else {
    return NULL;
  }
  while (!tree_nodes.empty()) {
    TreeNode* node = tree_nodes.front();
    tree_nodes.pop();
    bool left;
    cin >> left;
    if (left) {
      node->left = new TreeNode(nxt());
      tree_nodes.push(node->left);
    }
    bool right;
    cin >> right;
    if (right) {
      node->right = new TreeNode(nxt());
      tree_nodes.push(node->right);
    }
  }
  return root;
}
ListNode* makeLinkedList(int n) {
  ListNode* list = new ListNode(0);
  ListNode* dummy = list;
  for (int i = 0; i < n; ++i) {
    list->next = new ListNode(nxt());
    list = list->next;
  }
  return dummy->next;
}
const int N = 60;
array<int, N> operator^(const array<int, N>& a, const array<int, N>& b) {
  array<int, N> c;
  for (int i = 0; i < N; ++i) {
    c[i] = a[i] + b[i];
    if (c[i] >= 3) {
      c[i] -= 3;
    }
  }
  return c;
}
mt19937 rnd(42);
void solve() {
  int n = nxt();
  vector<int> in(n);
  for (int i = 0; i < n; ++i) {
    cin >> in[i];
    in[i]--;
  }
  vector<array<int, N>> nums(n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < N; ++j) {
      nums[i][j] = rnd() % 3;
    }
  }
  vector<array<int, N>> pr(1);
  map<array<int, N>, int> cnt;
  cnt[pr[0]] = 1;
  vector<vector<int>> pos(n);
  long long ans = 0;
  int cur = 0;
  for (int i = 0; i < n; ++i) {
    pr.push_back(pr.back() ^ nums[in[i]]);
    pos[in[i]].push_back(i);
    if (pos[in[i]].size() >= 4) {
      while (cur <= pos[in[i]][pos[in[i]].size() - 4]) {
        --cnt[pr[cur]];
        ++cur;
      }
    }
    ans += cnt[pr.back()];
    cnt[pr.back()]++;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(9);
  solve();
  return 0;
}
