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 main():
    input()
    return nPairsWithCommonFX(map(int, input().split()))


def nPairsWithCommonFX(sequence):
    storage = {}
    for value in sequence:
        y = f(value)
        if y not in storage:
            storage[y] = [value]
        else:
            storage[y].append(value)

    return sum(len(storage[y])*(len(storage[y])-1)//2 for y in storage)


def f(n):
    y = 1
    while n != 1:
        if n%2:
            y += 1
        n //= 2

    return y


if __name__ == '__main__':
    print(main())
    # main()
