Problem:
Score : 400 points

Problem Statement

We have N integers. The i-th integer is A_i.

Find \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

2 \leq N \leq 3 \times 10^5

0 \leq A_i < 2^{60}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the value \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

6

We have (1\mbox{ XOR } 2)+(1\mbox{ XOR } 3)+(2\mbox{ XOR } 3)=3+2+1=6.

Sample Input 2

10
3 1 4 1 5 9 2 6 5 3

Sample Output 2

237

Sample Input 3

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820

Sample Output 3

103715602

Print the sum modulo (10^9+7).

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val num = 64
    fun toBinaryNumberString(input: Long): String {
        return ((0 until num).map { "0" }.joinToString("") + java.lang.Long.toBinaryString(input)).takeLast(num)
    }

    val mod = 1000000007L
    val n = readLine()!!.toInt()
    val aList = readLine()!!.split(' ').map { toBinaryNumberString(it.toLong()).reversed() }
    val sumArray = Array(num) { 0L }
    (0 until num).forEach {
        for (i in 0 until n) {
            sumArray[it] += aList[i][it].toString().toLong()
        }
    }
    var ans = 0L
    (0 until num).forEach {
        ans += (((sumArray[it] * (n - sumArray[it])) % mod)
                * Math.pow(2.toDouble(), it.toDouble()).toLong() % mod) % mod
        ans %= mod
    }
    println(ans)
}