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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author tatsumack
 */

#include <iostream>
#include <fstream>

#include <iostream>
#include <limits.h>
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>

#include <list>
#include <map>
#include <numeric>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <unordered_map>
#include <unordered_set>

#define int long long
#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define REPS(i, n) for (int i = 1, i##_len = (n); i <= i##_len; ++i)
#define FOR(i, a, b) for (int i = (a), i##_len = (b); i <= i##_len; ++i)
#define REV(i, a, b) for (int i = (a); i >= (b); --i)
#define CLR(a, b) memset((a), (b), sizeof(a))
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define INF 1001001001001001001ll
#define fcout cout << fixed << setprecision(12)

using namespace std;

typedef pair<int, int> P;

int mod = 1e9 + 7;

struct mint {
    unsigned x;

    mint() : x(0) {}

    mint(signed sig) { x = sig < 0 ? sig % mod + mod : sig % mod; }

    mint(signed long long sig) { x = sig < 0 ? sig % mod + mod : sig % mod; }

    int get() const { return (int) x; }

    mint& operator+=(mint that) {
        if ((x += that.x) >= mod) x -= mod;
        return *this;
    }

    mint& operator-=(mint that) {
        if ((x += mod - that.x) >= mod) x -= mod;
        return *this;
    }

    mint& operator*=(mint that) {
        x = (unsigned long long) x * that.x % mod;
        return *this;
    }

    mint& operator/=(mint that) { return *this *= that.inverse(); }

    mint operator+(mint that) const { return mint(*this) += that; }

    mint operator-(mint that) const { return mint(*this) -= that; }

    mint operator*(mint that) const { return mint(*this) *= that; }

    mint operator/(mint that) const { return mint(*this) /= that; }

    mint inverse() const {
        long long a = x, b = mod, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b;
            std::swap(a, b);
            u -= t * v;
            std::swap(u, v);
        }
        return mint(u);
    }

    bool operator==(mint that) const { return x == that.x; }

    bool operator!=(mint that) const { return x != that.x; }

    mint operator-() const {
        mint t;
        t.x = x == 0 ? 0 : mod - x;
        return t;
    }
};

class EChangeALittleBit {
public:
    void solve(std::istream& cin, std::ostream& cout) {
        int N;
        cin >> N;
        vector<int> C(N);
        REP(i, N) cin >> C[i];

        sort(C.rbegin(), C.rend());

        vector<mint> two(N + 1);
        two[0] = 1;
        FOR(i, 1, N) two[i] = two[i - 1] * 2;

        mint res = 0;
        REP(i, N) {
            int r = i;
            int l = N - (r + 1);
            mint num = two[r];
            if (r >= 1) num += two[r - 1] * r;
            mint tmp = num * two[l] * C[i];
            res += tmp;
        }
        res *= two[N];
        cout << res.get() << endl;
    }
};


signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    std::istream& in(std::cin);
    std::ostream& out(std::cout);

    EChangeALittleBit solver;
    solver.solve(in, out);
    return 0;
}