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>

#define ii pair<int, int>
#define lli long long int
#define ll pair<lli, lli>
#define pb push_back
#define mp make_pair
#define fi first
#define se second

using namespace std;

lli bigmod = 1000000007;

lli fex(lli b) {
	lli ret;
	if (b == 0)
		return 1;
	ret = fex(b / 2);
	ret = (ret * ret) % bigmod;
	if (b & 1)
		ret = (ret * 2) % bigmod;
	return ret;
}

int main() {
	lli n, c[200005], ans = 0, temp;
	scanf("%lld", &n);
	for (lli i = 0; i < n; i++)
		scanf("%lld", &c[i]);
	sort(c, c + n);
	for (lli i = 0; i < n; i++) {
		temp = ((n - i + 1) * c[i]) % bigmod;
		ans = (ans + temp) % bigmod;
	}
	ans = (ans * fex(n + n - 2)) % bigmod;
	printf("%lld\n", ans);
	return 0;
}