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;
typedef long long ll;

ll calc(ll a,ll b,ll p){//a^(b) mod p   
	if(b==0)return 1;
	ll y = calc(a,b/2,p);y=(y*y)%p;
	if(b & 1) y = (y * a) % p;
	return y;
}


int main(){
	ll N,MOD=1000000007;
	cin >> N;
	vector<ll> C(N);
	for(auto &x:C) cin >> x;

	sort(C.rbegin(), C.rend());
	
	ll ans = 0;
	ll tr = calc(2,MOD-2,MOD);
	ll fn = calc(4,N-1,MOD);

	for(int i=0;i<N;i++){
		( ans += (i+2)*fn%MOD*C[i] ) %= MOD;
	}

	cout << ans << endl;
}

