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>
#define rep(i, n) for(int i=0; i<(n); i++)
#define ALL(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
const int MAX = 300002;
const int mod = 1e9 + 7;

int main(){
	ll N; cin >> N;
	ll C[N];
	for(int i=0; i<N; i++) cin >> C[i];
	sort(C, C+N);
	ll t[2*N];
	t[0] = 1;
	for(int i=1; i<2*N; i++) t[i] = t[i-1]*2%mod;
	ll ans = 0;
	for(int i=0; i<N; i++){
		if(i != N-1)
		ans = (ans + ((t[2*N-2]*(N-i+1))%mod)*C[i])%mod;
		else
		ans = (ans+C[i]*t[2*i+1]%mod)%mod;
	}
	cout << ans << endl;
	return 0;
}
