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<cstdio>
#include<algorithm>

using namespace std;

using ll = long long;
const int mod = 1e9+7;
int a[(int)2e5 + 10];

void add(int &x,int y){
    x += y;
    x = x >= mod ? x - mod : x;
}
int subt(int x, int y){
    x -=y;
    x = x < 0 ? x + mod : x;
    return x;
}
int mult(int x,int y){
    return (1LL * x* y) %mod;
}
int main(){
    int n;
    scanf("%d",&n);
    for(int i= 1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    sort(a+1,a+n+1);
    reverse(a+1,a+n+1);
    int ans = 0,mul = 2,bi = 1;
    for(int i = 1;i<=n;i++){
        ans = mult(ans,4);
        add(ans , mult(mul, a[i]));
        bi = mult(4, bi);
        mul = mult(bi, i + 2);
    }
    printf("%d",ans);
}
