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;
const long long MOD = 1e9+7;
int main(){
    int n;
    cin >> n;
    long long c[n];
    rep(i,n) cin >> c[i];
    sort(c,c+n);
    long long nibeki[2*n];
    nibeki[0]=1;
    rep(i,2*n-1){
        nibeki[i+1]=2*nibeki[i];
        nibeki[i+1]%=MOD;
    } 
    long long ans = 0;
    rep(i,n){
        long long tmp = c[i];
        tmp%=MOD;
        long long rank = n-i;
        long long tmp2 = (rank-1)*nibeki[rank-2]+nibeki[rank-1];
        tmp2%=MOD;
        tmp*=tmp2;
        tmp%=MOD;
        tmp*=nibeki[2*n-rank];
        tmp%=MOD;
        ans+=tmp;
        ans%=MOD;
    }
    cout << ans << endl;
}