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 int long long
#define F first
#define S second

const int MAXM = 6 * 100 * 1000;
const int MOD = 1000 * 1000 * 1000 + 7;
int tav[MAXM], a[MAXM];

int32_t main() {
	ios::sync_with_stdio(false);
	tav[0] = 1;
	for (int i = 1; i < MAXM; i++) {
		tav[i] = tav[i - 1] * 2 % MOD;
	}
	int n, ans = 0;
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	sort(a, a + n);
	for (int i = 0; i < n; i++)
		ans = (ans + a[i] * tav[2 * n - 2] % MOD * (n - i + 1) % MOD) % MOD;
//		cerr << ans << endl;
	cout << ans;
}

