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).

Examples

Input

1
1000000000


Output

999999993


Input

2
5 8


Output

124


Input

5
52 67 72 25 79


Output

269312
#include "bits/stdc++.h"
using namespace std;

long long MOD = 1000 * 1000 * 1000 + 7;

long long p(long long a, long long b) {
	if (b == 0) return 1;
	if (b == 1) return a;
	long long ha = p(a, b/2);
	if (b % 2 == 0) return (ha * ha) % MOD;
	return (((ha * ha) % MOD) * a) % MOD;
}

int main() {
	
	long long n;
	cin >> n;
	long long ar[n];
	for (int i=0; i<n; i++) cin >> ar[i];
	
	sort(ar, ar+n);
	reverse(ar, ar+n);
	
	long long x = p(2, 2*n-2);
	long long s = 0;
	for (int i=1; i<=n; i++) {
		s += (i + 1) * (ar[i-1]);
		s %= MOD;
	}
	
	cout << (x * s) % MOD << endl;
	
	
}
