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;
const ll MOD = 1e9+7;
const int N = 2e5;

ll dmod(ll x){
	return x%MOD;
}

ll fe(ll x, ll p){
	if(p==0) return 1ll;
	if(p==1) return x;
	ll ret = fe(x,p/2);
	ret = dmod(ret*ret);
	if(p&1) ret = dmod(ret*x);
	return ret;
}

int n;
ll li[N+5];

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld",&li[i]);
	}
	sort(li+1,li+n+1);
	ll ans = 0ll;
	for(int i=1;i<=n;i++){
		ans = dmod(ans + dmod(li[i] * dmod(fe(2ll,i-1) * dmod(fe(2ll,n-i) + dmod((ll)(n-i) * fe(2ll,n-i-1))	))));
//		cout << dmod(fe(2ll,n-i) + dmod((ll)(n-i) * fe(2ll,n-i-1))) << endl;;
	}
	ans = dmod(ans * fe(2ll,n));
	printf("%lld\n",ans);
	return 0;
}
