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 <algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep1(i, n) for (int i = 1; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const int INF = 1001001001;

const int mod = 1000000007;
struct mint {
  ll x;

  mint(ll x=0):x((x%mod+mod)%mod){}

  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

int main() {
  int N;
  cin >> N;
  vector<ll> C(N);
  rep(i,N) cin >> C[i];

  sort(C.begin(), C.end(), greater<>());  
  mint sum = 0;
  mint base = 1;
  rep(i,N-1) base *= 2;
  rep(i,N) {    
    mint d = base + mint(i)*base/2;
    mint c = C[i];
    sum += c * d;
  }
  mint n = 1;
  rep(i,N) n *= 2;
  mint ans = sum*n;  
  cout << ans.x << endl;
  return 0;
}