Vasya has a sequence a consisting of n integers a_1, a_2, ..., a_n. Vasya may pefrom the following operation: choose some number from the sequence and swap any pair of bits in its binary representation. For example, Vasya can transform number 6 (... 00000000110_2) into 3 (... 00000000011_2), 12 (... 000000001100_2), 1026 (... 10000000010_2) and many others. Vasya can use this operation any (possibly zero) number of times on any number from the sequence.

Vasya names a sequence as good one, if, using operation mentioned above, he can obtain the sequence with [bitwise exclusive or](https://en.wikipedia.org/wiki/Exclusive_or) of all elements equal to 0.

For the given sequence a_1, a_2, …, a_n Vasya'd like to calculate number of integer pairs (l, r) such that 1 ≤ l ≤ r ≤ n and sequence a_l, a_{l + 1}, ..., a_r is good.

Input

The first line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — length of the sequence.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^{18}) — the sequence a.

Output

Print one integer — the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and the sequence a_l, a_{l + 1}, ..., a_r is good.

Examples

Input

3
6 7 14


Output

2


Input

4
1 2 1 16


Output

4

Note

In the first example pairs (2, 3) and (1, 3) are valid. Pair (2, 3) is valid since a_2 = 7 → 11, a_3 = 14 → 11 and 11 ⊕ 11 = 0, where ⊕ — bitwise exclusive or. Pair (1, 3) is valid since a_1 = 6 → 3, a_2 = 7 → 13, a_3 = 14 → 14 and 3 ⊕ 13 ⊕ 14 = 0.

In the second example pairs (1, 2), (2, 3), (3, 4) and (1, 4) are valid.
#include <bits/stdc++.h>
using namespace std;
long long int get_num_one_bits(long long int num) {
  long long int ans = 0;
  while (num != 0) {
    ans += (num % 2);
    num = num / 2;
  }
  return ans;
}
int main() {
  long long int n;
  cin >> n;
  vector<long long int> a(n + 1), b(n + 1), pref_sum(n + 1);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++) b[i] = get_num_one_bits(a[i]);
  for (int i = 1; i <= n; i++) pref_sum[i] = pref_sum[i - 1] + b[i];
  vector<long long int> even_inds, odd_inds;
  even_inds.clear();
  odd_inds.clear();
  for (int i = 0; i <= n; i++) {
    if (pref_sum[i] % 2 == 0)
      even_inds.push_back(i);
    else
      odd_inds.push_back(i);
  }
  long long int ans = 0;
  vector<long long int> inds;
  for (int p = 0; p <= 1; p++) {
    if (p == 0)
      inds = even_inds;
    else
      inds = odd_inds;
    for (int l = 0; l < (long long int)inds.size() - 1; l++) {
      int j = 1;
      if (inds[l + 1] - inds[l] == 1) j++;
      while (((l + j) < inds.size()) && (inds[l + j] - inds[l] <= 64)) {
        int flag = 1;
        long long int sum = pref_sum[inds[l + j]] - pref_sum[inds[l]];
        for (int k = inds[l] + 1; k <= inds[l + j]; k++) {
          if (b[k] > sum / 2) {
            flag = 0;
            break;
          }
        }
        if (flag == 1) ans++;
        j++;
      }
      long long int curr = l + j;
      ans += (inds.size() - 1) - curr + 1;
    }
  }
  cout << ans << endl;
}
