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;
using ll = long long;
using ull = unsigned long long;
#define REP(i,n) for (ll i = 0; i < (n); ++i)

struct mint{
    static const long long MOD = 1000000007;

    long long v;

    mint() : v(0) {}
    mint(long long x){
        x = x % MOD;
        if(x < 0){ x += MOD; }
        v = x;
    }
    mint& operator+=(mint a){
        v += a.v;
        if(v >= MOD){ v -= MOD; }
        return *this;
    }
    mint& operator*=(mint a){
        v *= a.v;
        v %= MOD;
        return *this;
    }
    mint& operator-=(mint a){
        v -= a.v;
        if(v < 0){ v += MOD; }
        return *this;
    }
    mint& operator/=(mint a){
        return (*this) *= a.inv();
    }

    mint operator*(mint a) const{
        return mint(v) *= a;
    }

    mint operator+(mint a) const {
        return mint(v) += a;
    }

    mint operator-(mint a) const {
        return mint(v) -= a;
    }

    mint operator/(mint a) const {
        return mint(v) /= a;
    }

    mint pow(long long k) const {
        mint res(1),tmp(v);
        while(k){
            if(k&1) res *= tmp;
            tmp *= tmp;
            k >>= 1;
        }
        return res;
    }
    mint inv(){return pow(MOD - 2); }

    static mint comb(long long n, int k){
        mint res(1);
        for(int i = 0; i < k; ++i){
            res *= mint(n - i);
            res /= mint(i + 1);
        }
        return res;
    }

    static mint factorial(long long n){
        mint res(1);
        for(int i = n; i > 1; --i){
            res *= mint(i);
        }
        return res;
    }

    bool operator<(const mint &a) const {
        return v < a.v;
    };
};

mint operator*(long long l, mint r){ return r * l; }

std::ostream& operator<<(std::ostream& stream, const mint& m){
    stream << m.v;
    return stream;
}

void solve(long long N, std::vector<long long> C){
    sort(C.begin(), C.end());
    vector<mint> mC(N);
    REP(i, N){ mC.at(i) = mint(C.at(i)); }

    vector<mint> p2(N + 1);
    p2.at(0) = 1;
    REP(i, N){ p2.at(i + 1) = 2 * p2.at(i); }

    mint ans = 0;
    REP(i, N){
        // C_i を含む項の和を考える
        mint tmp = p2.at(N) * mC.at(i) * p2.at(i);
        mint tmp2 = 1;
        const ll j = N - i;
        tmp2 = p2.at(j - 1) + (j - 1) * p2.at(max(0LL, j - 2));
        ans += tmp * tmp2;
    }
    cout << ans << endl;
}

// Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools
int main(){
    long long N;
    scanf("%lld",&N);
    std::vector<long long> C(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%lld",&C[i]);
    }
    solve(N, std::move(C));
    return 0;
}