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: Accepted)

private const val MOD = 1e9.toLong() + 7

fun main(args: Array<String>) {
    val N = readLine()!!.toInt()
    val A = readLine()!!.split(" ").map {it.toLong()}

    var res = ModInt(0L, MOD)
    var p = ModInt(1L, MOD)

    for (i in 0 until 60) {
        val cnt0 = A.count { it and (1L shl i) == 0L }.toLong()
        res += (p * cnt0 * (N - cnt0))
        p *= 2L
    }

    println(res.value)
}

class ModInt (value: Long, val mod: Long) {
    init {
        if (mod <= 0) {
            throw IllegalArgumentException("ModInt cannot be created for non-positive modulus: $mod")
        }
    }

    val value = (value % mod + mod) % mod

    private fun assertSameMod(other: ModInt) {
        if (mod != other.mod) {
            throw IllegalArgumentException(
                "You cannot operate summation, subtraction and multiplication for ModInts with different moduli: $mod, ${other.mod}"
            )
        }
    }

    operator fun unaryMinus() = ModInt(mod - value, mod)

    operator fun plus(other: ModInt): ModInt {
        assertSameMod(other)

        return ModInt((value + other.value) % mod, mod)
    }

    operator fun plus(other: Long) = plus(ModInt(other, mod))

    operator fun minus(other: ModInt): ModInt {
        assertSameMod(other)

        return plus(-other)
    }

    operator fun minus(other: Long) = minus(ModInt(other, mod))

    operator fun times(other: ModInt): ModInt {
        assertSameMod(other)

        return ModInt((value * other.value) % mod, mod)
    }

    operator fun times(other: Long) = times(ModInt(other, mod))

    fun pow(k: Long): ModInt {
        var res = 1L
        var a = value
        var l = k

        while (l > 0) {
            if ((l and 1L) == 1L) {
                res = (res * a) % mod
            }

            a = a * a % mod
            l /= 2L
        }

        return ModInt(res, mod)
    }

    /**
     * Returns an inverse element of the value.
     * To ensure `m * m.inv() = 1`, the modulus should be prime, since it uses Fermat's little theorem.
     */
    fun inv() = pow(mod - 2)

    operator fun div(other: ModInt): ModInt {
        assertSameMod(other)
        if (other.value == 0L) {
            throw ArithmeticException()
        }

        return times(other.inv())
    }

    operator fun div(other: Long) = div(ModInt(other, mod))
}

class ModCombination(val n: Long, val mod: Long) {
    private val fact = Array(n.toInt() + 1) { ModInt(1, mod) }
    private val invfact = Array(n.toInt() + 1) { ModInt(1, mod) }

    init {
        for (i in 2..n.toInt()) {
            fact[i] = fact[i - 1] * i.toLong()
            invfact[i] = fact[i].inv()
        }
    }

    operator fun get(n: Long, m: Long): ModInt {
        if (m > n) {
            return ModInt(0, mod)
        }

        return fact[n.toInt()] * invfact[m.toInt()] * invfact[(n - m).toInt()]
    }

    fun calc(n: Long, m: Long): ModInt {
        return fact[(n * m).toInt()] * invfact[n.toInt()].pow(m) * invfact[m.toInt()]
    }
}