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 REP(i, n)  for (int i = 0; i < (int) (n); i++)
#define REP1(i, n) for (int i = 1; i <= (int) (n); i++)
#define all(x)     (x).begin(), (x).end()
#define SZ(x)      ((int) (x).size())
using ll = long long;
const ll mod = 1e9 + 7;

int main() {
  int n;
  cin >> n;
  vector <ll> c(n);
  REP(i, n) {
	cin >> c[i];
  }
  sort(all(c), greater <ll>());
  if (n == 1) {
	  cout << 2 * c[0] % mod << endl;
	  return 0;
	}
  ll t1 = 1;
  ll t2 = 1;
  REP(i, 2 * n - 1) {
	t1 = t1 * 2 % mod;
  }
  REP(i, 2 * n - 2) {
	t2 = t2 * 2 % mod;
  }
  ll ans = 0;
  REP(i, n) {
	ans = (ans + ((t1 + t2 * i) % mod) * c[i] % mod) % mod;
  }
  cout << ans << endl;
  return 0;
}
