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 <vector>
#include <algorithm>
using namespace std;
const int MOD = 1000000007;

long long mod(long long a) {
	a %= MOD;
	return a >= 0 ? a : a+MOD;
}

long long pow_mod(long long a, long long n) {
	long long res = 1;
	while (n > 0) {
		if (n & 1) res = (res * a) % MOD;
		n >>= 1;
		a = (a * a) % MOD;
	}
	return res;
}

int main() {
	long long n;
	cin >> n;
	vector<long long> C(n);
	for (int i = 0; i < n; i++) cin >> C[i];
	sort(C.begin(), C.end());
	long long ans = 0;
	for (int i = 0; i < n; i++) {
		ans = mod(ans + C[n-1-i] * (2 + i) % MOD);
	}
	ans = mod(ans * pow_mod(2, (n-1)<<1));
	cout << ans << endl;
	return 0;
}