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: Accepted)



#include<bits/stdc++.h>
using namespace std;
#define inf INT_MAX
#define INF LLONG_MAX
#define ll long long
#define ull unsigned long long
#define M (int)(1e9+7)
#define P pair<int,int>
#define FOR(i,m,n) for(int i=(int)m;i<(int)n;i++)
#define RFOR(i,m,n) for(int i=(int)m;i>=(int)n;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,n,0)
#define all(a) a.begin(),a.end()
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define int ll
#define vi vector<int>
#define IP pair<int,P>
#define PI pair<P,int>
#define PP pair<P,P>
#define Yes(f){cout<<(f?"Yes":"No")<<endl;}
#define YES(f){cout<<(f?"YES":"NO")<<endl;}



struct Mint {
  int x;

  Mint() : x(0) {}

  Mint(int y) : x(y >= 0 ? y % M : (M - (-y) % M) % M) {}

  Mint &operator+=(const Mint &p) {
    if((x += p.x) >= M) x -= M;
    return *this;
  }

  Mint &operator-=(const Mint &p) {
    if((x += M - p.x) >= M) x -= M;
    return *this;
  }

  Mint &operator*=(const Mint &p) {
    x = (int) (1LL * x * p.x % M);
    return *this;
  }

  Mint &operator/=(const Mint &p) {
    *this *= p.inverse();
    return *this;
  }

  Mint operator-() const { return Mint(-x); }

  Mint operator+(const Mint &p) const { return Mint(*this) += p; }

  Mint operator-(const Mint &p) const { return Mint(*this) -= p; }

  Mint operator*(const Mint &p) const { return Mint(*this) *= p; }

  Mint operator/(const Mint &p) const { return Mint(*this) /= p; }

  bool operator==(const Mint &p) const { return x == p.x; }

  bool operator!=(const Mint &p) const { return x != p.x; }

  Mint inverse() const {
    int a = x, b = M, u = 1, v = 0, t;
    while(b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return Mint(u);
  }

  Mint power(int n) const {
    Mint ret(1), mul(x);
    while(n > 0) {
      if(n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const Mint &p) {
    return os << p.x;
  }

  friend istream &operator>>(istream &is, Mint &a) {
    int t;
    is >> t;
    a = Mint(t);
    return (is);
  }


};
using mint = Mint;



template< typename T>
T power(T x,int k){
  T ret=1;
  while(k>0){
    if(k&1) (ret*=x);
    (x*=x); 
    k >>= 1;
  }
  return ret;
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(20);

  int n;
  cin>>n;
  vi v(n);
  rep(i,n){
    cin>>v[i];
  }
  sort(all(v));

  Mint ans=0;
  int m=2*(n-1);
  Mint mol = power<Mint>(2,m);
  Mint ml=mol;
  mol*=2;
  rrep(i,n-1){
    Mint w=v[i];
    //cout<<mol<<endl;
    ans+=w*mol;
    mol+=ml;
    
  }

  cout<<ans<<endl;


}