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 <fstream>
#include <algorithm>       // for next_permutation
#include <vector>
#include <queue>
#include <bits/stl_numeric.h>
#include <math.h>

#define DEBUG 0
#define REP(i, n) for (long long i = 0; i < (n); i++) 
using namespace std;

typedef long long ll;
typedef pair <int, int> pii;
static const ll MOD = 1000000007;
static const ll INF = 1000000000000000000LL;

long myPow(long x, long n, long m){
  if(n == 0)
    return 1;
  if(n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}

int main() {

    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N;
    cin >> N;

    vector <ll> c(N);

    REP(i, N)
    {
        cin >> c[i];
    }
    sort(c.rbegin(), c.rend());

    ll tmp3 = myPow(2, N * 2 - 2, MOD);
    ll tmp2 = tmp3 * 2 % MOD;
    ll res = 0;

    REP(i,N)
    {
        ll tmp;
        tmp = tmp2 * c[i] % MOD;

        res += tmp;
        res %= MOD;

        tmp2 += tmp3;
        tmp2 %= MOD;
    }

    cout << res << endl;

    return 0;
}