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;
typedef long long int ll;
ll c[200005],p[200005],hell=pow(10,9)+7;
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll n,res=0;
    cin>>n;
    p[0]=1;
    for(ll i=1;i<200005;i++)
    p[i]=(2*p[i-1])%hell;
    for(ll i=0;i<n;i++)
    cin>>c[i];
    sort(c,c+n);
    for(ll i=0;i<n;i++){
        ll sum=(p[n-1]+((n-i-1)*p[n-2])%hell)%hell;
        sum*=c[i];
        sum%=hell;
        res+=sum;
        res%=hell;
    }
    res*=p[n];
    res%=hell;
    cout<<res;
}
