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).

Examples

Input

1
1000000000


Output

999999993


Input

2
5 8


Output

124


Input

5
52 67 72 25 79


Output

269312
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define endl '\n'
#define D(x) cerr << #x << " = " << (x) << '\n'
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()

const int MOD = 1e9 + 7;
const int N = 2e5 + 5;

int n;
int arr[N];

int main()
{
  int n;
  cin >> n;
  for(int i = 0 ; i < n ; i++)
    cin >> arr[i];
  sort(arr, arr + n, greater<int>());
  long long ans = 0;
  for(int i = 0 ; i < n ; i++){
    ans += (1LL * arr[i] * (i + 2)) % MOD;
    ans %= MOD;
  }
  for(int i = 1 ; i < n ; i++)
    ans = (ans * 4) % MOD;
  cout << ans << endl;
  return 0;
}
