Dima got into number sequences. Now he's got sequence *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>, consisting of *n* positive integers. Also, Dima has got a function *f*(*x*), which can be defined with the following recurrence:
 -  *f*(0) = 0; -  *f*(2·*x*) = *f*(*x*); -  *f*(2·*x* + 1) = *f*(*x*) + 1. 
Dima wonders, how many pairs of indexes (*i*, *j*) (1 ≤ *i* &lt; *j* ≤ *n*) are there, such that *f*(*a*<sub class="lower-index">*i*</sub>) = *f*(*a*<sub class="lower-index">*j*</sub>). Help him, count the number of such pairs.
The first line contains integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>). The second line contains *n* positive integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>).

The numbers in the lines are separated by single spaces.
def f(x):
    return bin(x).count("1")


from math import comb


def solve():
    # Reading input
    n = int(input())
    a = list(map(int, input().split()))

    freq = {}
    for ai in a:
        ai = f(ai)
        freq[ai] = freq.get(ai, 0) + 1

    count = sum([comb(f, 2) for f in freq.values()])
    print(count)


solve()
