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 mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long a[300005];
long long ev[300005];
long long odd[300005];
long long sum[300005];
int main() {
  ios::sync_with_stdio(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long p;
    cin >> p;
    a[i] = __builtin_popcountll(p);
  }
  for (long long i = n - 1; i >= 0; i--) {
    sum[i] = sum[i + 1] + a[i];
    odd[i] += odd[i + 1];
    ev[i] += ev[i + 1];
    if (sum[i] & 1) {
      odd[i - 1]++;
    } else {
      ev[i - 1]++;
    }
  }
  long long ans = 0;
  for (long long i = n - 2; i >= 0; i--) {
    if (sum[i] & 1) {
      ans += odd[i];
    } else {
      ans += ev[i];
      ans++;
    }
    long long sm = 0;
    long long maxx = 0;
    for (long long j = i; j < min((long long)100 + i, n); j++) {
      sm += a[j];
      maxx = max(maxx, a[j]);
      if (sm % 2 == 0 && 2 * maxx > sm) {
        ans--;
      }
    }
  }
  cout << ans << "\n";
}
