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<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
const int mod = 1e9 + 7;
const int N = 4e5 + 11;
int n;
LL c[N], mi[N];
inline int M(int x, int y){
    return 1LL * x * y % mod;
}
inline int A(int x, int y){
    return x + y - (x + y >= mod ? mod : 0);
}
int main(){
    cin>>n;
    mi[0] = 1;
    for(int i = 1;i <= n; i++){
        scanf("%lld", &c[i]);
        mi[i] = M(mi[i-1], 2);
    }
    for(int i = n + 1;i <= 2 * n; i++)mi[i] = M(2, mi[i-1]);
    sort(c + 1, c + 1 + n);
    int ans = 0;
    for(int i = 1;i <= n; i++){
        ans = A(ans, M(M(mi[n+i-1], c[i]), A(M(n - i, mi[n-i-1]), mi[n-i])));
    }
    cout<<ans<<endl;
    return 0;
}
