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;
# define rep(i,n) for(int i=0;i<(n);i++)
typedef long long ll;
const ll MOD = 1000000007;

// aのn乗のMOD
ll modpow(ll a,ll n){
    ll res = 1;
    while(n>0){
        // &１つはビット積
        if(n&1) res = res * a%MOD;
        a = a*a%MOD;
        n>>=1;
    }
    return res;
}

int main(){
	cin.tie(0);
   	ios::sync_with_stdio(false);
	
	ll N;
	cin >> N;
    vector<ll> C(N);
	rep(i,N){
		cin>>C[i];
	}
	sort(C.begin(),C.end(),greater<ll>());
	ll ans =0;
	ll s = modpow(2,2*N-2);
	
	rep(i,N){
		ans =(ans+ (((s*C[i])%MOD)*(i+2)))%MOD;
	}
	cout<<ans<<"\n";
	return 0;
}
