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;
using lint = long long;
#define low lower_bound
#define upp upper_bound
#define rep(i, n) for (lint i = 0; i < lint(n); ++i)
#define Rep(i, n) for (lint i = lint(n); i >= 0; --i)
#define all(vec) (vec).begin(), (vec).end()
#define cend printf("\n");
#define prique(T) priority_queue<T, vector<T>, greater<T>>
#define prique2(T) priority_queue<T>

struct edge { lint to; lint cost; };
using P = pair<lint, lint>;
using D = pair<double, double>;
//const lint mod = 998244353LL;
const lint mod = 1000000007LL;
const lint inf = 3LL * mod * mod;
const double Pi = acos(-1.0);

template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmin(T& a, T b) {
	return chmax(b, a);
}

lint modpow(lint x, lint n, lint m) {
	lint ans = 1;
	while (n > 0) {
		if (n & 1) {
			ans *= x;
			if (m <= ans) ans %= m;
		}
		x *= x;
		if (m <= x) x %= m;
		n >>= 1;
	}
	return ans;
}

lint fact[1000000];
void f_init(lint n) {
	if (1000000 <= n) return;
	fact[0] = fact[1] = 1;
	for (lint i = 2; i <= n; ++i) {
		fact[i] = i * fact[i - 1];
		if (mod <= fact[i]) fact[i] %= mod;
	}
	return;
}

lint inver[1000000]; bool inve[100000];
lint inv(lint x) {
	if (x == 0) return 0;
	else if (inve[x]) return inver[x];
	else {
		inve[x] = true;
		return inver[x] = modpow(x, mod - 2, mod);
	}
}

lint comb(lint n, lint r) {
	if (n < r) return 0;
	if (n == r) return 1;
	lint ans = fact[n] * inv(fact[n - r]) % mod * inv(fact[r]) % mod;
	if (ans < 0) return ans + mod;
	return ans;
}

lint nck[1010][1010];
void comb(lint n) {
	rep(i, n + 1) {
		rep(j, i + 1) {
			if (j == 0 || i == j) nck[i][j] = 1;
			else nck[i][j] = nck[i - 1][j - 1] + nck[i - 1][j];
		}
	}
}

lint gcd(lint a, lint b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}

lint lcm(lint a, lint b) { return a / gcd(a, b) * b; }

bool pri(lint p) {
	for (lint i = 2; i * i <= p; i == 2 ? ++i : i += 2) if (p % i == 0) return false;
	return p > 1;
}

map<lint, lint> factring(lint n) {
	map<lint, lint>ans;
	for (lint i = 2; i * i <= n; ++i) {
		while (n % i == 0) {
			n /= i;
			++ans[i];
		}
	}
	if (n != 1) ++ans[n];
	return ans;
}

lint xor128() {
	static lint x = 123456789, y = 362436069, z = 521288629, w = 88675123;
	lint t = (x xor (x << 11));
	x = y; y = z; z = w;
	return (w = (w xor (w >> 19)) xor (t xor (t >> 8)));
}

double dis(D a, D b) {
	double x = a.first - b.first, y = a.second - b.second;
	return sqrt(x * x + y * y);
}

class unionfind {
private:
	int par[2 * 100010], ranks[2 * 100010], sizeth[2 * 100010];
public:
	void init(int n) {
		for (int i = 0; i < n; ++i) {
			par[i] = i;
			ranks[i] = 0;
			sizeth[i] = 1;
		}
	}

	int root(int x) {
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	bool same(int x, int y) {
		return root(x) == root(y);
	}

	void unite(int x, int y) {
		x = root(x);
		y = root(y);
		if (x == y) return;
		if (ranks[x] < ranks[y]) {
			par[x] = y;
			sizeth[y] += sizeth[x];
		}
		else {
			par[y] = x;
			if (ranks[x] == ranks[y]) ++ranks[x];
			sizeth[x] += sizeth[y];
		}
	}
};


lint n, m, k, ans;
lint c[2 * 100010];
int main() {
	cin >> n;
	rep(i, n) {
		cin >> c[i];
	}
	sort(c, c + n, greater<lint>());
	m = modpow(2, n - 1, mod);
	k = modpow(2, n - 2, mod);
	rep(i, n) {
		ans += (m + i * k) % mod * c[i];
		ans %= mod;
	}
	cout << ans * modpow(2, n, mod) % mod << endl;
}