Problem:
Score : 500 points

Problem Statement

For two sequences S and T of length N consisting of 0 and 1, let us define f(S, T) as follows:

Consider repeating the following operation on S so that S will be equal to T. f(S, T) is the minimum possible total cost of those operations.

Change S_i (from 0 to 1 or vice versa). The cost of this operation is D \times C_i, where D is the number of integers j such that S_j \neq T_j (1 \leq j \leq N) just before this change.

There are 2^N \times (2^N - 1) pairs (S, T) of different sequences of length N consisting of 0 and 1. Compute the sum of f(S, T) over all of those pairs, modulo (10^9+7).

Constraints

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

1 \leq C_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
C_1 C_2 \cdots C_N

Output

Print the sum of f(S, T), modulo (10^9+7).

Sample Input 1

1
1000000000

Sample Output 1

999999993

There are two pairs (S, T) of different sequences of length 2 consisting of 0 and 1, as follows:

S = (0), T = (1): by changing S_1 to 1, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

S = (1), T = (0): by changing S_1 to 0, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

The sum of these is 2000000000, and we should print it modulo (10^9+7), that is, 999999993.

Sample Input 2

2
5 8

Sample Output 2

124

There are 12 pairs (S, T) of different sequences of length 3 consisting of 0 and 1, which include:

S = (0, 1), T = (1, 0)

In this case, if we first change S_1 to 1 then change S_2 to 0, the total cost is 5 \times 2 + 8 = 18. We cannot have S = T at a smaller cost, so f(S, T) = 18.

Sample Input 3

5
52 67 72 25 79

Sample Output 3

269312

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val N = readLine()!!.toInt()
    val C = readLine()!!.split(" ").map{it.toLong()}.sorted()
    val MOD = 1e9.toLong() + 7L

    var res = ModInt(0L, MOD)
    val TWO = ModInt(2L, MOD)

    for (i in C.indices) {
        val m = N - i - 1L
        res += TWO.pow(i.toLong()).pow(2L) * (TWO.pow(m - 1L) * m + TWO.pow(m)) * TWO.pow(m) * C[i] * 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()]
    }
}