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 rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

const ll MOD = 1e9 + 7;

ll pow(ll a, ll b){
	if(b == 0) return 1;
	if(b % 2 == 1) return a * pow(a, b - 1) % MOD;

	ll d = pow(a, b / 2) % MOD;
	return d * d % MOD;
}

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	ll n;
	cin >> n;

	vector<ll> c(n);
	rep(i, n) cin >> c[i];
	sort(c.begin(), c.end());
	if(n == 1){
		cout << 2 * c[0] % MOD << endl;
		return 0;
	}

	ll ans = pow(2ll, n);

	ll x = 0;
	for(ll i = 0; i < n; i++){
		ll y = c[i];
		y *= pow(2ll, n - 2);
		y %= MOD;
		y *= n - i + 1;
		y %= MOD;
		
		x += y;
		x %= MOD;
	}

	ans *= x;
	ans %= MOD;

	cout << ans << endl;
}
