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 <algorithm>
#include <cmath>
#include <complex>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
#include <functional>
#include <cstring>
#include <regex>
#include <random>
#include <cassert>
using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, s, n) for (int i = (s); i < (int)(n); i++)
#define revrep(i, n) for (int i = (n) - 1; i >= 0; i--)
#define revrepr(i, s, n) for (int i = (n) - 1; i >= s; i--)
#define debug(x) cerr << #x << ": " << x << "\n"

using ll = long long;
using P = pair<int, int>;

const long long MOD = 1e9+7;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class T>
istream& operator >>(istream &is, vector<T> &v) {
    for (int i = 0; i < (int)v.size(); i++) cin >> v.at(i);
    return is;
}

template<class T, class U>
ostream& operator <<(ostream &os, pair<T, U> p) {
    cout << '(' << p.first << ", " << p.second << ')';
    return os;
}

template<class T>
void print(const vector<T> &v, const string &delimiter) { rep(i, v.size()) cout << (0 < i ? delimiter : "") << v.at(i); cout << endl; }

template<class T>
void print(const vector<vector<T>> &vv, const string &delimiter) { for (const auto &v: vv) print(v, delimiter); }

struct ModInt {
    static ll mod;
    ll val;
    ModInt(ll v = 0) : val(v % mod) {}
    ModInt operator-() const { return ModInt(val ? mod - val : val); }
    ModInt &operator+=(ModInt a) {
        if ((val += a.val) >= mod) val -= mod;
        return *this;
    }
    ModInt &operator-=(ModInt a) {
        if ((val -= a.val) < 0) val += mod;
        return *this;
    }
    ModInt &operator*=(ModInt a) {
        val = (__uint128_t(val) * a.val) % mod;
        return *this;
    }
    ModInt &operator/=(ModInt a) {
        ll u = 1, v = a.val, s = 0, t = mod;
        while (v) {
            ll q = t / v;
            swap(s -= u * q, u);
            swap(t -= v * q, v);
        }
        a.val = (s < 0 ? s + mod : s);
        val /= t;
        return (*this) *= a;
    }
    ModInt inv() const { return ModInt(1) /= (*this); }
    bool operator<(ModInt x) const { return val < x.val; }
};
ll ModInt::mod = 1e9 + 7;

ostream &operator<<(ostream &os, ModInt a) {
    os << a.val;
    return os;
}
ModInt operator+(ModInt a, ModInt b) { return a += b; }
ModInt operator-(ModInt a, ModInt b) { return a -= b; }
ModInt operator*(ModInt a, ModInt b) { return a *= b; }
ModInt operator/(ModInt a, ModInt b) { return a /= b; }
ModInt pow(ModInt a, ll e) {
    ModInt x(1);
    for (; e > 0; e /= 2) {
        if (e % 2 == 1) x *= a;
        a *= a;
    }
    return x;
}

ModInt comb(int n, int k) {
    ModInt x = 1, y = 1;
    rep(i, k) {
        x *= n - i;
        y *= i + 1;
    }
    return x / y;
}

int main() {
    int n;
    cin >> n;
    vector<ll> c(n);
    cin >> c;
    sort(c.begin(), c.end());
    
    ModInt ans;
    rep(i, n) {
        ModInt a = pow(ModInt(2), n + i);
        ModInt b = pow(ModInt(2), n - i - 1) + pow(ModInt(2), (n - i - 2)) * (n - i - 1);
        ans += c[i] * a * b;
    }
    cout << ans << endl;
}