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;
typedef long long ll;
const int N=2e5+5,mod=1e9+7;
int n;
ll a[N],f[N];
int main()
{
    f[0]=1;
    for(int i=1;i<N;i++)
        f[i]=f[i-1]*4%mod;
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    sort(a+1,a+1+n);
    ll ans=0;
    for(int i=1;i<=n;i++)
    {
        ans=(ans+a[i]*2*f[n-1]%mod)%mod;
        if(n-i-1>=0)
            ans=(ans+a[i]*4%mod*(n-i)%mod*f[n-i-1]%mod*f[i-1]%mod)%mod;
    }
    printf("%lld\n",ans);
}
