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 <iostream>
#include <cstdio>
#include <algorithm>
#define N 200005
using namespace std;

typedef long long ll;
ll M = 1e9 + 7;
ll n, s, ans, po[N]{1}, c[N];
int main()
{
	ll i;
	cin >> n;
	for (i = 1; i <= n; i++) {
		po[i] = po[i - 1] * 2 % M;
	}
	for (i = 0; i < n; i++) scanf("%lld", &c[i]);
	sort(c, c + n);
	s = po[n - 1];
	for (i = n - 1; i >= 0; i--) {
		ans = (ans + s * c[i]) % M;
		s = (s + po[n - 2]) % M;
	}
    cout << ans * po[n] % M << endl;
    return 0;
}