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 <bits/stdc++.h>

using namespace std;

template <class T>inline T updmax(T& a, T b) {return a = max(a, b);}
template <class T>inline T updmin(T& a, T b) {return a = min(a, b);}

template <unsigned long long mod> class modint {
public:
	unsigned long long v;
	modint(const long long x = 0) : v(x % mod) {}
	modint operator+(const modint rhs) { return modint(*this) += rhs; }
	modint operator-(const modint rhs) { return modint(*this) -= rhs; }
	modint operator*(const modint rhs) { return modint(*this) *= rhs; }
	modint operator/(const modint rhs) { return modint(*this) /= rhs; }
	modint operator-() { return modint(mod - this->v); }
	modint& operator+=(const modint rhs) {
		v += rhs.v;
		if (v >= mod) v -= mod;
		return *this;
	}
	modint& operator-=(const modint rhs) {
		if (v < rhs.v) v += mod;
		v -= rhs.v;
		return *this;
	}
	modint& operator*=(const modint rhs) {
		v = v * rhs.v % mod;
		return *this;
	}
	modint inverse(modint a) {
		unsigned long long exp = mod - 2;
		modint ret(1ULL);
		while (exp) {
			if (exp % 2) { ret *= a; }
			a *= a;
			exp >>= 1;
		}
		return ret;
	}
	modint& operator/=(modint rhs) {
		(*this) *= inverse(rhs);
		return *this;
	}
	friend ostream& operator<<(ostream& os, modint& u) {
		os << u.v;
		return (os);
	}
	friend istream& operator>>(istream& is, modint& u) {
		is >> u.v;
		return (is);
	}
};

const int MOD = 1000000007;
using mint = modint<MOD>;
template<class T>
T fastpow(T a, long long p){
	T tmp = a;
	T ret = 1;
	while(p){
		if(p & 1)ret *= tmp;
		p >>= 1;
		tmp = tmp * tmp;
	}
	return ret;
}

class Solution {
public:
	void solve() {
		int n;
		cin >> n;
		vector<int> a(n, 0);
		for(int i=0; i<n; i++)cin >> a[i];
		sort(a.begin(), a.end(), greater<int>());
		mint ans = 0;
		for(int i=0; i < n; i++)ans += mint{a[i]} * (i+2);
		ans *= fastpow(mint{2}, 2*n-2);
		cout << ans << endl;
		return;
	};
};

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	Solution solution;
	solution.solve();
	return 0;
}