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 <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep_r(i, n) for (int i = n - 1; i >= 0; i--)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define bit(n) (1LL << (n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

#define INF bit(60)
#define pb push_back
#define mod 1000000007

using namespace std;
using uif = uint_fast64_t;
using ll = long long int;
using tTree = __gnu_pbds::tree<ll, __gnu_pbds::null_type, less<ll>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;

ll dx[4] = {1, 0, -1, 0};
ll dy[4] = {0, 1, 0, -1};

#define FACSIZE 200002

ll invfac[FACSIZE];
ll fac[FACSIZE];

template <class T>
bool chmax(T &, const T &);
template <class T>
bool chmin(T &, const T &);
ll gcd(ll, ll);
ll powLL(ll x, ll y);

ll mod_pow(ll, ll);
ll mod_add(ll, ll);
ll mod_mul(ll, ll);
ll mod_div(ll, ll);
ll comb(ll, ll);
void make_fact(ll);
void make_invfact(ll);
void fact_init();

int main(void)
{
  ll n;
  cin >> n;
  vector<ll> C(n);
  rep(i, n) cin >> C[i];
  sort(begin(C), end(C));

  make_fact(FACSIZE - 1);
  make_invfact(FACSIZE - 1);

  ll ans = 0;

  for (ll i = 0; i < n; i++)
  {
    ll con = mod_mul(C[i], mod_pow(2LL, n));
    con = mod_mul(con, mod_pow(2LL, i));
    /*
    for (ll k = 0; k <= n - 1 - i; k++)
    {
      ll val = mod_mul(k + 1, mod_mul(con, comb(n - 1 - i, k)));
      //cout << i << "," << k << endl;
      //cout << val << endl;
      ans = mod_add(ans, val);
    }
    */
    ll alpha = n - 1 - i;
    ll val = 0;
    if (alpha == 0)
    {
      val = mod_mul(0 + 1, mod_mul(con, comb(n - 1 - i, 0)));
    }
    else
    {
      val = mod_mul(alpha + 2, mod_pow(2LL, alpha - 1));
      val = mod_mul(val, con);
    }

    //cout << val << endl;
    ans = mod_add(ans, val);
  }

  cout << ans << endl;

  return 0;
}

ll mod_pow(ll x, ll r)
{
  if (r == 0)
    return 1;
  else if (r == 1)
    return x % mod;
  else if (r % 2 == 0)
  {
    ll t = mod_pow(x, r / 2) % mod;
    return mod_mul(t, t);
  }
  else
  {
    ll t = mod_pow(x, r / 2) % mod;
    ll k = mod_mul(t, t);
    return (k % mod) * (x % mod);
  }
}
ll mod_add(ll a, ll b)
{
  return ((a % mod) + (b % mod)) % mod;
}

ll mod_mul(ll a, ll b)
{
  return ((a % mod) * (b % mod)) % mod;
}

ll mod_div(ll a, ll b)
{
  return mod_mul(a, mod_pow(b, mod - 2));
}

void fact_init()
{
  make_fact(FACSIZE - 1);
  make_invfact(FACSIZE);
}

void make_fact(ll n)
{
  fac[0] = 1;
  rep(i, n)
  {
    fac[i + 1] = mod_mul(fac[i], i + 1);
  }
}

void make_invfact(ll n)
{
  invfac[n] = mod_pow(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--)
  {
    invfac[i] = mod_mul(invfac[i + 1], i + 1);
  }
}

ll comb(ll n, ll r)
{

  return mod_mul(mod_mul(fac[n], invfac[r]), invfac[n - r]);
}
template <class T>
bool chmax(T &a, const T &b)
{
  if (a < b)
  {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
  if (b < a)
  {
    a = b;
    return 1;
  }
  return 0;
}

ll qp(ll a, ll b)
{
  ll ans = 1LL;
  do
  {
    if (b & 1LL)
      ans = 1LL * mod_mul(ans, a) % mod;
    a = 1LL * mod_mul(a, a) % mod;
  } while (b >>= 1LL);
  return ans;
}
ll qp(ll a, ll b, ll mo)
{
  ll ans = 1LL;
  do
  {
    if (b & 1LL)
      ans = 1LL * (ans % mo) * (a % mo);
    a = 1LL * (a % mo) * (a % mo);
  } while (b >>= 1LL);
  return ans;
}
ll gcd(ll a, ll b)
{
  return b ? gcd(b, a % b) : a;
}
ll powLL(ll x, ll y)
{
  ll ans = 1LL;
  for (ll i = 0LL; i < y; i++)
    ans *= x;
  return ans;
}