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;
const int N=2e5+50,mod=1e9+7;
int n,c[N],ans,pw[N];
int read(){
    int x=0,c;
    while(!isdigit(c=getchar()));
    while(isdigit(c))x=x*10+c-48,c=getchar();
    return x;
}
int main(){
    n=read();pw[0]=1;
    for(int i=1;i<=n;i++)pw[i]=pw[i-1]*2%mod;
    for(int i=1;i<=n;i++)c[i]=read();
    sort(c+1,c+n+1,[&](int a,int b){
        return a>b;
    });
    for(int i=1;i<=n;i++){
        int dat=pw[i-1];
        if(i>1)dat=(dat+1ll*pw[i-2]*(i-1))%mod;
        ans=(ans+1ll*dat*c[i]%mod*pw[n-i])%mod;
    }
    printf("%d\n",1ll*ans*pw[n]%mod);
    return 0;
}