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;
typedef long long ll;
ll mod = 1e9+7;
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	ll n;
	cin>>n;
	ll a[n+5];
	for(ll i=0;i<n;i++)
	{
		cin>>a[i];
	}
	sort(a,a+n);
	ll p[n+5];
	p[0]=1;
	for(ll i=1;i<=n;i++)
	{
		p[i]=p[i-1]*2;
		if (p[i]>=mod)
			p[i]%=mod;
	}

	ll ans = 0;
	for(ll i=1;i<=n;i++)
	{
		ll Ans = p[n-i-1]*(n-i);
		if (Ans>=mod)
			Ans%=mod;
		Ans += p[n-i];
		if (Ans>=mod)
			Ans%=mod;

		Ans*=p[i-1];
		if (Ans>=mod)
			Ans%=mod;

		Ans*=a[i-1];
		if (Ans>=mod)
			Ans%=mod;
		ans+=Ans;
		if (ans>=mod)
			ans%=mod;
		//cout<<Ans<<" ";
	}
	//cout<<"\n";
	ans*=p[n];
	if (ans>=mod)
		ans%=mod;
	cout<<ans;
}