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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

#define Mod 1000000007

long long pow2(long long k)
{
	static long long ans[200001] = {1, 2};
	if (ans[k] != 0) return ans[k];
	
	long long i;
	for (i = 1; i * 2 <= k; i *= 2) ans[i*2] = ans[i] * ans[i] % Mod;
	ans[k] = ans[i] * pow2(k - i) % Mod;
	return ans[k];
}

void merge_sort(int x[], int n)
{
	static int y[200001] = {};
	if (n <= 1) return;
	merge_sort(&(x[0]), n/2);
	merge_sort(&(x[n/2]), (n+1)/2);
	
	int i, p, q;
	for (i = 0, p = 0, q = n/2; i < n; i++) {
		if (p >= n/2) y[i] = x[q++];
		else if (q >= n) y[i] = x[p++];
		else y[i] = (x[p] < x[q])? x[p++]: x[q++];
	}
	for (i = 0; i < n; i++) x[i] = y[i];
}

int main()
{
	int i, N, C[200001];
	scanf("%d", &N);
	for (i = 0; i < N; i++) scanf("%d", &(C[i]));
	merge_sort(C, N);
	if (N == 1) {
		printf("%d\n", C[0] * 2 % Mod);
		fflush(stdout);
		return 0;
	}
	
	long long ans = 0;
	for (i = N - 1; i >= 0; i--) ans = (ans + pow2(N - 2) * (N - i + 1) % Mod * C[i]) % Mod;
	printf("%lld\n", ans * pow2(N) % Mod);
	fflush(stdout);
	return 0;
}