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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cassert>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <numeric>
#include <algorithm>

using namespace std;
using lint = long long;
constexpr int MOD = 1000000007, INF = 1010101010;
constexpr lint LINF = 1LL << 60;

template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
	for (const auto &e : vec) os << e << (&e == &vec.back() ? "\n" : " ");
	return os;
}

#ifdef _DEBUG
template <class T>
void dump(const char* str, T &&h) { cerr << str << " = " << h << "\n"; };
template <class Head, class... Tail>
void dump(const char* str, Head &&h, Tail &&... t) {
	while (*str != ',') cerr << *str++; cerr << " = " << h << "\n";
	dump(str + (*(str + 1) == ' ' ? 2 : 1), t...);
}
#define DMP(...) dump(#__VA_ARGS__, __VA_ARGS__)
#else 
#define DMP(...) ((void)0)
#endif

template<int Modulo = MOD>
struct Mint {

	lint val;
	constexpr Mint(lint v = 0) noexcept : val(v % Modulo) { if (val < 0) val += Modulo; }

	constexpr Mint& operator += (const Mint &r) noexcept {
		val += r.val;
		if (val >= Modulo) val -= Modulo;
		return *this;
	}
	constexpr Mint& operator -= (const Mint &r) noexcept {
		val -= r.val;
		if (val < 0) val += Modulo;
		return *this;
	}
	constexpr Mint& operator *= (const Mint &r) noexcept {
		val = val * r.val % Modulo;
		return *this;
	}
	constexpr Mint& operator /= (const Mint &r) noexcept {
		lint a = r.val, b = Modulo, u = 1, v = 0;
		while (b) {
			lint t = a / b;
			a -= t * b; swap(a, b);
			u -= t * v; swap(u, v);
		}
		val = val * u % Modulo;
		if (val < 0) val += Modulo;
		return *this;
	}

	constexpr Mint operator + (const Mint &r) const noexcept { return Mint(*this) += r; }
	constexpr Mint operator - (const Mint &r) const noexcept { return Mint(*this) -= r; }
	constexpr Mint operator * (const Mint &r) const noexcept { return Mint(*this) *= r; }
	constexpr Mint operator / (const Mint &r) const noexcept { return Mint(*this) /= r; }

	constexpr int getmod() { return Modulo; }
	constexpr Mint operator - () const noexcept { return val ? Modulo - val : 0; }

	constexpr bool operator == (const Mint &r) const noexcept { return val == r.val; }
	constexpr bool operator != (const Mint &r) const noexcept { return val != r.val; }

	friend ostream &operator << (ostream &os, const Mint<Modulo> &x) noexcept { return os << x.val; }
	friend istream &operator >> (istream &is, Mint<Modulo> &x) noexcept {
		lint tmp; is >> tmp;
		x = Mint<Modulo>(tmp);
		return is;
	}
	friend constexpr Mint<Modulo> modpow(const Mint<Modulo> &a, lint n) noexcept {
		Mint res(1), tmp = a;
		while (n > 0) {
			if (n & 1) res *= tmp;
			tmp *= tmp;
			n >>= 1;
		}
		return res;
	}
};

using mint = Mint<>;

lint modpow(lint a, lint n, lint mod = MOD) {
	lint res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int N;
	cin >> N;

	vector<int> c(N);
	for (int i = 0; i < N; i++) cin >> c[i];
	sort(c.begin(), c.end());

	mint ans = 0;
	for (int i = 0; i < N; i++) {
		ans += modpow(mint(4), N - 1) * (N - i + 1) * c[i];
	}

	cout << ans << "\n";

	return 0;
}