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<algorithm>
#include<vector>
#include<string>
#include<utility>
#include<map>
#include<set>
#include<queue>
#include<math.h>
using namespace std;
#define N (1000000000+7)
#define INF 1e16
typedef long long ll;
typedef pair<ll,ll> P;


int main(void){
    ll n;
    cin>>n;
    vector<ll>c(n);
    for(ll i=0;i<n;i++)cin>>c[i];
    sort(c.begin(),c.end());
    ll ans=1;
    for(ll i=0;i<2*n-2;i++){
        ans = ans*2;
        ans = ans%N;
    }
    ll ans2 = 0;
    for(ll i=0;i<n;i++){
        ll tmp =(n-i+1)*c[i];
        tmp = tmp%N;
        ans2 = (ans2+tmp)%N;
    }
    cout<<(N+(ans*ans2)%N)%N<<endl;
    return 0;
}