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"
#include<cassert>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
const int mod = 1000000007;
const int inf = 1ll << 61;
typedef pair<int, int> P;
typedef pair<int,P>PP;

int modpow(int a, int x, int mod) {
	int res = 1;
	while (x) {
		if (x & 1)res = res*a%mod;
		a = a*a%mod;
		x >>= 1;
	}
	return res;
}
int c[200006];
int d[200006];
signed main() {
	int n; cin >> n;
	rep(i, n)cin >> c[i];
	sort(c, c + n); reverse(c, c + n);
	d[0] = modpow(2, n - 1, mod);
	rep(i, n - 1)d[i + 1] = (d[i] + modpow(2, n - 2, mod)) % mod;
	int res = 0;
	rep(i, n)res = (res + c[i] * d[i]) % mod;
	res = res*modpow(2, n, mod) % mod;
	cout << res << endl;
}