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 <iostream>
#include <iomanip>
#include <string>
#include <stack>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <iterator>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <regex>
#include <complex>
using namespace std;
using pii  = pair<int,int>;
using ll=long long;
using ld=long double;
#define pb push_back
#define mp make_pair
#define sc second
#define fr first
#define stpr setprecision
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define rep(i,n) for(ll i=0;i<(n);++i)
#define Rep(i,a,b) for(ll i=(a);i<(b);++i)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define rRep(i,a,b) for(int i=a;i>=b;i--)
#define crep(i) for(char i='a';i<='z';++i)
#define psortsecond(A,N) sort(A,A+N,[](const pii &a, const pii &b){return a.second<b.second;});
#define ALL(x) (x).begin(),(x).end()
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define endl '\n'
int ctoi(const char c){
  if('0' <= c && c <= '9') return (c-'0');
  return -1;
}
ll gcd(ll a,ll b){return (b == 0 ? a : gcd(b, a%b));}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
constexpr ll MOD=1000000007;
constexpr ll INF=1000000011;
constexpr ll MOD2=998244353;
constexpr ll LINF = 1001002003004005006ll;
constexpr ld EPS=10e-8;
template <class T, class U> inline bool chmax(T& lhs, const U& rhs) { if (lhs < rhs) { lhs = rhs; return 1; } return 0; }
template <class T, class U> inline bool chmin(T& lhs, const U& rhs) { if (lhs > rhs) { lhs = rhs; return 1; } return 0; }
template<typename T> istream& operator>>(istream& is,vector<T>& v){for(auto&& x:v)is >> x;return is;}
template<typename T,typename U> istream& operator>>(istream& is, pair<T,U>& p){ is >> p.first; is >> p.second; return is;}
template<typename T,typename U> ostream& operator>>(ostream& os, const pair<T,U>& p){ os << p.first << ' ' << p.second; return os;}
template<class T> ostream& operator<<(ostream& os, vector<T>& v){
  for(auto i=begin(v); i != end(v); ++i){
    if(i !=begin(v)) os << ' ';
    os << *i;
  }
  return os;
}

ll cme;

ll modpow(ll b, ll e, ll m){  // bのe乗のmod m を返す
  if (e <= 0) return 1;
  cme = modpow(b, e / 2, m);
  cme = (cme * cme) % m;
  if (e % 2 == 1) cme = (cme * b) % m;
  return cme;
}

int main(){
  ll N,A[200007];
  cin >> N ;
  ll H = modpow(2,N,MOD),K = modpow(2,N-1,MOD), U = modpow(2,N-2,MOD);
  rep(i,N){
    cin >> A[i];
  }
  sort(A,A+N);
  reverse(A,A+N);
  ll C=0,X=K;
  rep(i,N){
    C+=X*A[i];
    C%=MOD;
    X+=U;
    X%=MOD;
  }
  cout << C*H%MOD << endl;
}