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 <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll dp[200000],n,a[200000],b[200000],MOD=1000000007,ans;
ll f(ll x,ll y){
    if(y<=0)return 1;
    if(y%2==1)return f(x,y-1)*x%MOD;
    return f(x,y/2)*f(x,y/2)%MOD;
}
int main(void){
    cin>>n;
    for(int i=0;i<n;i++)cin>>dp[i];
    sort(dp,dp+n);
    for(ll i=0LL;i<n;i++){
        a[i]=f(2,i);
        b[n-i-1]=(f(2,i)+i*f(2,i-1)%MOD)%MOD;
    }
    for(int i=0;i<n;i++){
        ans+=dp[i]*a[i]%MOD*b[i]%MOD;
        ans%=MOD;
    }
    cout<<ans*f(2,n)%MOD<<endl;
}
