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;
vector<long long> a, pre, ev;
vector<vector<long long>> nxt;
long long n;
long long c2(long long n) { return n * (n - 1) / 2; }
int32_t main() {
  cin >> n;
  a.resize(n);
  pre.assign(n + 1, 0);
  ev.assign(2, 0);
  ev[0] = 1;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    a[i] = __builtin_popcountll(a[i]);
    pre[i + 1] = pre[i] + a[i];
    ev[pre[i + 1] % 2]++;
  }
  nxt.resize(n, vector<long long>(64, n));
  long long cnt = c2(ev[1]) + c2(ev[0]);
  for (long long l = 0; l < n; ++l) {
    long long su = 0, mx = 0;
    for (long long r = l; r < min(n, l + 125); ++r) {
      su += a[r];
      mx = max(mx, a[r]);
      if (su < 2LL * mx && su % 2 == 0) cnt--;
    }
  }
  cout << cnt << endl;
  return 0;
}
