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)

// #define DEBUGGING
#include <bits/stdc++.h>
#define endl '\n'
#define ALL(V) (V).begin(), (V).end()
#define ALLR(V) (V).rbegin(), (V).rend()
using ll = std::int64_t;
using ull = std::uint64_t;
using PLL = std::pair<ll, ll>;
using TLL = std::tuple<ll, ll, ll>;
template <typename T> using V = std::vector<T>;
template <typename T> using VV = V<V<T>>;
template <typename T> const T& var_min(const T &t) { return t; }
template <typename T> const T& var_max(const T &t) { return t; }
template <typename T, typename... Tail> const T& var_min(const T &t, const Tail&... tail) { return std::min(t, var_min(tail...)); }
template <typename T, typename... Tail> const T& var_max(const T &t, const Tail&... tail) { return std::max(t, var_max(tail...)); }
template <typename T, typename... Tail> void chmin(T &t, const Tail&... tail) { t = var_min(t, tail...); }
template <typename T, typename... Tail> void chmax(T &t, const Tail&... tail) { t = var_max(t, tail...); }
template <typename T> const T& clamp(const T &t, const T &low, const T &high) { return std::max(low, std::min(high, t)); }
template <typename T> void chclamp(T &t, const T &low, const T &high) { return t = clamp(t, low, high); }
namespace init__ { struct InitIO { InitIO() { std::cin.tie(nullptr); std::ios_base::sync_with_stdio(false); std::cout << std::fixed << std::setprecision(30); } } init_io; }
#define mv_rec make_v(init, tail...)
template <typename T> T make_v(T init) { return init; }
template <typename T, typename... Tail> auto make_v(T init, size_t s, Tail... tail) { return V<decltype(mv_rec)>(s, mv_rec); }
#undef mv_rec
using namespace std;

#ifdef DEBUGGING
#include "../../debug/debug.cpp"
#else
#define DEBUG(...) 0
#define DEBUG_SEPARATOR_LINE 0
#endif


const ll mod = 1e9 + 7;;

ll mod_pow(ll n, ll k) {
    if (n == 0) return 1;
    ll ret = 1, cur = n;
    while (k) {
        if (k & 1) (ret *= cur) %= mod;
        k /= 2;
        cur = cur * cur % mod;
    }
    return ret;
}

ll mod_inv(ll n) { return mod_pow(n, mod - 2); }

int main() {
    ll N;
    cin >> N;
    V<ll> C(N);
    for (ll &e : C) cin >> e;
    ll inv2 = mod_inv(2);
    ll inv4 = inv2 * inv2 % mod;
    ll pow4 = 1;
    ll ans = 0;
    for (ll i = 0; i < N; i++) (pow4 *= 4) %= mod;
    ll val = pow4 * inv4 % mod;
    ll len = N - 1;
    sort(ALL(C));
    for (ll e : C) {
        ans += e * (len * val % mod) % mod;
        ans += (e * val % mod * 2 % mod); 
        len--;
    }
    cout << ans % mod << endl;
    return 0;
}