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 <iostream>
#include <algorithm>
#include <string>
#define llint long long
#define mod 1000000007

using namespace std;

llint n;
llint c[200005];

llint modpow(llint a, llint n)
{
	if(n == 0) return 1;
	if(n % 2){
		return ((a%mod) * (modpow(a, n-1)%mod)) % mod;
	}
	else{
		return modpow((a*a)%mod, n/2) % mod;
	}
}


int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> c[i];
	sort(c+1, c+n+1);
	reverse(c+1, c+n+1);
	
	
	llint inv4 = modpow(4, mod-2);
	
	llint ans = 0;
	for(int i = 1; i <= n; i++){
		ans += inv4 * (i+1) % mod * c[i] % mod;
		ans %= mod;
	}
	ans *= modpow(2, 2*n), ans %= mod;
	cout << ans << endl;
	
	return 0;
}