Problem:
Score : 500 points

Problem Statement

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).

Sample Input 1

1
1000000000

Sample Output 1

999999993

There are two pairs (S, T) of different sequences of length 2 consisting of 0 and 1, as follows:

S = (0), T = (1): by changing S_1 to 1, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

S = (1), T = (0): by changing S_1 to 0, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

The sum of these is 2000000000, and we should print it modulo (10^9+7), that is, 999999993.

Sample Input 2

2
5 8

Sample Output 2

124

There are 12 pairs (S, T) of different sequences of length 3 consisting of 0 and 1, which include:

S = (0, 1), T = (1, 0)

In this case, if we first change S_1 to 1 then change S_2 to 0, the total cost is 5 \times 2 + 8 = 18. We cannot have S = T at a smaller cost, so f(S, T) = 18.

Sample Input 3

5
52 67 72 25 79

Sample Output 3

269312

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("Ofast")
#include <cstdio>
#include <cstdlib>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
#include <bitset>
#include <cmath>
#include <limits>
#include <iostream>
#include <map>
#include <set>
#include <tuple>
using namespace std;
#define INF 1LL << 61
#define MAX 1LL << 30
#define MOD 1000000007
typedef long long ll;
typedef pair<int,int> P;
//typedef pair<pair<int,int>,int> p;
#define bit(n, k) ((n >> k) & 1) /*nのk bit目*/
#define rad_to_deg(rad) (((rad) / 2 / M_PI) * 360)
struct edge{ll to,cost,val;};
template<class T,class U>bool chmin(T&a,const U&b){if(a<=b)return false;a=b;return true;}
template<class T,class U>bool chmax(T&a,const U&b){if(a>=b)return false;a=b;return true;}
//__builtin_popcount(S);
//C.erase(unique(C.begin(),C.end()),C.end());
//#define int ll
int dx[]={1,-1,0,0},dy[]={0,0,-1,1};

signed main(void)
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    cin>>N;
    ll C[210000];
    for(int i=0;i<N;i++) cin>>C[i];

    sort(C,C+N,greater<ll>());
    for(int i=N-1;i>=1;i--){
        C[i-1]+=C[i];
    }
    ll t[420000];
    t[0]=1;
    for(int i=1;i<=2*N;i++){
        t[i]=t[i-1]*2;
        t[i]%=MOD;
    }
    ll ans=0;
    for(int i=0;i<N;i++){
        //cout<<t[2*N-1-i]<<" "<<C[i]<<endl;
        ans+=t[2*N-1-i]*C[i];
        ans%=MOD;
    }
    cout<<ans<<endl;

}