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 <bits/stdc++.h>
#include <math.h>
#define _GLIBCXX_DEBUG
#define _LIBCPP_DEBUG 0

using namespace std;
#define ll long long
#define rep(i,n) for (int i = 0; i < n; i++)
#define rrep(i,n) for (int i = n-1; i >= 0; i--)
#define MOD (1000000007)
#define vi vector<int>
#define vl vector<ll>
#define vb vector<bool>
#define vvi vector<vi>
#define vvl vector<vl>
#define pii pair<int, int>
#define pli pair<ll, int>a
#define pb push_back
#define sz(x) int(x.size())

#define mp make_pair
#define all(a) (a).begin(),(a).end()

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

ll gcd(ll a, ll b) {
    if (b == 0) return a;
    else return gcd(b, a % b);
}

ll keta(ll n){
  string s = to_string(n);
  ll num = s.size();
  return num;
}

const ll INF = 1LL << 60;

const int dh[4] = {1,0,-1,0};
const int dw[4] = {0,1,0,-1};

struct Edge{
  int to;
  int weight;
  Edge(int t, int w) : to(t), weight(w){}
};
using Graph = vector<vector<Edge>>;
using P = pair<ll, int>;

// dist[v] に頂点 s から頂点 v への距離を入れる関数
void dijkstra(int s, Graph &G, vector<ll> &dist){
  fill(all(dist), INF);
  dist[s] = 0;

  priority_queue<P, vector<P>, greater<P>> pq;//距離が小さい順に取り出す
  pq.push(P(0,s)); // 頂点 s からスタート

  while (!pq.empty()) {
    P p = pq.top(); pq.pop();
    int v = p.second;
    if(dist[v] < p.first) continue;// 既に v に辿り着く短い経路有ならスルー (これはただの高速化)

    for(auto e : G[v]){// v から行ける頂点 nv 達を走査
      int nv = e.to;
      ll w = e.weight;
      if(dist[nv] > dist[v] + w){
        dist[nv] = dist[v] + w;
        pq.push(P(dist[nv], nv));
      }
    }
  }
}

class UnionFind{
public:
  vi Parent;
  UnionFind(int n){
    Parent = vi(n,-1);
  }

  int root(int a){
    if(Parent[a] < 0) return a;
    else return Parent[a] = root(Parent[a]);
  }

  int size(int a){
    return -Parent[root(a)];
  }

  bool merge(int a, int b){
    a = root(a);
    b = root(b);
    if(a == b) return false;

    if(size(a) < size(b)) swap(a,b);
    Parent[a] += Parent[b];
    Parent[b] = a;

    return true;
  }
};

class Factrial{
public:
  vl Fac;
  Factrial(int MAX){
    Fac = vl(MAX+1);
    rep(i,MAX) {
      if(i == 0) Fac[i] = 1;
      else Fac[i+1] = ((i+1)*Fac[i])%MOD;
    }
  }
};



int main(){
  int n; cin >> n;
  vl c(n); rep(i,n) cin >> c[i];
  sort(all(c));

  vl t(n+1,1);
  rep(i,n) t[i+1] = (t[i] * 2)%MOD;

  ll ans = 0;
  rep(i,n) {
    int l = i, r = n-i-1;
    ll tmp = t[r];
    if(r != 0) tmp = (t[r] + t[r-1] * r)%MOD;
    tmp = (tmp * t[l])%MOD;
    tmp = (tmp * c[i])%MOD;
    ans = (ans + tmp)%MOD;
  }
  ans = (ans * t[n])%MOD;
  cout << ans << endl;
}