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 <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
#define ll long long
#define ld long double
using namespace std;

ll C[200010], mod=1000000007;

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