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
//abc150_e.cpp
//Sat Feb 29 10:37:18 2020

#include <iostream>
#include <string>
#include <queue>
#include <map>
#include <unordered_map>
#include <vector>
#include <algorithm>
#include <math.h>
#include <set>
#define INTINF 2147483647
#define LLINF 9223372036854775807
using namespace std;
using ll=long long;
typedef pair<int,int> P;

int main(){
	ll n;
	ll MOD = 1000000007;
	cin >> n;

	ll c[n];
	for (int i=0;i<n;i++){
		cin >> c[i];
	}

	sort(c,c+n,greater<ll>());

	ll ans = 0;
	for (ll i=0;i<n;i++){
		ll tmp = (i+2)*c[i]%MOD;
		ans = (ans+tmp)%MOD;
	}

	for (ll i=0;i<n-1;i++){
		ans = ans*4%MOD;
	}
	cout << ans << endl;
}