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 rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
constexpr int MX = 200005, MOD = 1e9+7;
int n;
ll c[MX], ans;
ll exp(ll ret, ll n){
	for(ll b = 2; n; n/=2,(b*=b)%=MOD) if(n%2) (ret *= b) %= MOD;
	return ret;
}
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("%lld", c+i);
	sort(c,c+n,greater<ll>());
	rep(i,n) (ans += c[i] * (i+2)) %= MOD;
	printf("%lld\n", exp(ans,n*2-2));
}