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 <vector>
#include <algorithm>
using ll=long long;
const int mod=1e9+7;
long long modpow(long long a, long long n) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
int main(){
    int n;
    std::cin>>n;
    std::vector<ll> c(n);
    for(int i=0;i<n;i++)std::cin>>c[i];
    ll ret=0;
    std::sort(c.begin(),c.end());
    for(int i=0;i<n;i++){
        if(n-i-1>0)ret+=c[i]*modpow(2,i*2)%mod*2%mod*modpow(2,2*(n-i-1)-1)%mod*((n-i-1)+2)%mod;
        else ret+=c[i]*modpow(2,i*2)%mod*2%mod;
        ret%=mod;
        //std::cout<<ret<<std::endl;
    }
    std::cout<<ret<<std::endl;
}