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 <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <cassert>
#include <cfloat>
#include <complex>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

using namespace std;

#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define Lower_bound(v, x) \
  distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) \
  distance(v.begin(), upper_bound(v.begin(), v.end(), x))

using ll = long long;
using ull = unsigned long long;
using P = pair<ll, ll>;
using T = tuple<ll, ll, ll>;
using vll = vector<ll>;
using vP = vector<P>;
using vT = vector<T>;
using vvll = vector<vector<ll>>;
using vvP = vector<vector<P>>;
using dqll = deque<ll>;

ll dx[9] = {-1, 1, 0, 0, -1, -1, 1, 1, 0};
ll dy[9] = {0, 0, -1, 1, -1, 1, -1, 1, 0};

const ll INF = 1LL << 50;

const ll mod = 1000000007;

ll fact[200200];
ll invfact[200200];

inline ll take_mod(ll a) { return (a % mod + mod) % mod; }

inline ll add(ll a, ll b) { return take_mod(a + b); }

inline ll sub(ll a, ll b) { return take_mod(a - b); }

inline ll mul(ll a, ll b) { return take_mod(a * b); }

inline ll pow(ll x, ll n) {
  ll res = 1LL;
  while (n > 0) {
    if (n & 1) res = mul(res, x);
    x = mul(x, x);
    n >>= 1;
  }
  return res;
}

ll mod_inv(ll x) { return pow(x, mod - 2); }

void make_fact(ll n) {
  fact[0] = 1;
  ll res = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = res;
    res = mul(res, i + 1);
  }
}

void make_invfact(ll n) {
  invfact[0] = 1;
  invfact[n] = mod_inv(fact[n]);
  for (int i = n - 1; i >= 1; i--) {
    invfact[i] = mul(invfact[i + 1], i + 1);
  }
}

ll perm(ll n, ll k) { return mul(fact[n], invfact[n - k]); }

ll comb(ll n, ll k) { return mul(mul(fact[n], invfact[n - k]), invfact[k]); }

int main() {
  ll n;
  cin >> n;

  vll c(n);
  rep(i, n) cin >> c[i];
  Sort(c);
  ll ans = 0;

  ll tt[n + 1];
  tt[0] = 1;
  rep(i, n) tt[i + 1] = mul(tt[i], 2);

  rep(i, n) {
    ll addv = tt[n - i - 1];
    if (i != n - 1) addv = add(addv, mul(n - i - 1, tt[n - i - 2]));
    addv = mul(addv, tt[i]);
    addv = mul(addv, c[i]);
    ans = add(ans, addv);
  }
  ans = mul(ans, tt[n]);

  cout << ans << endl;

  return 0;
}