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 rp(i,n) for(ll i=0; i<n; i++)
using namespace std;
using ll=long long;

ll RepeatSquare(ll x,ll n,ll mod){
    if(n==0) return 1;
    if(n==1) return x%mod;
    if(n%2==0){
        ll half=RepeatSquare(x,n/2,mod);
        return (half*half)%mod;
    }
    if(n%2==1){
        return x*RepeatSquare(x,n-1,mod)%mod;
    }
}

int main(){
    ll n;
    cin >> n;
    vector<ll> c(n);
    rp(i,n) cin >> c.at(i);
    sort(c.begin(),c.end());
    reverse(c.begin(),c.end());
    ll ans=0;
    rp(i,n){
        ans += c.at(i)*(i+2);
        ans %= 1000000007;
    }
    ans*=RepeatSquare(4,n-1,1000000007);
    ans%=1000000007;
     cout << ans << endl;
     return 0;
}