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
import java.util.*;
import java.io.*;

class Main{
    public static void main(String[] args){
        FastScanner fsc=new FastScanner();
        int n=fsc.nextInt();
        long ans=0;
        long[] c=new long[n];
        long mod=1_000_000_007;
        for(int i=0;i<n;i++){
            c[i]=fsc.nextLong();
        }
        Arrays.sort(c);
        if(n==1){
            System.out.println((c[0]*2)%mod);
            return;
        }
        for(int i=0;i<n;i++){
            ans+=c[i]*(n-i+1);
            ans%=mod;
        }
        fsc.close();
        ans*=pow2(2*n-2, mod);
        ans%=mod;
        System.out.println(ans);
    }
    static long pow2(long n, long mod){
        if(n==0) return 1;
        if(n==1) return 2;
        return (pow2(n/2, mod)*pow2(n-n/2, mod))%mod;
    }
    static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;
        private boolean hasNextByte(){
            if(ptr<buflen) return true;
            else{
                ptr = 0;
                try{
                    buflen = in.read(buffer);
                }
                catch(IOException e){
                    e.printStackTrace();
                }
                if(buflen<=0) return false;
            }
        return true;
        }
        private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
        private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
        public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
        public String next() {
            if (!hasNext()) throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while(isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }
        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) throw new NumberFormatException();
            while(true){
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                }
                else if(b == -1 || !isPrintableChar(b)) return minus ? -n : n;
                else throw new NumberFormatException();
                b = readByte();
            }
        }
        public int nextInt() {
            long nl = nextLong();
            if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
            return (int) nl;
        }
        public double nextDouble() { return Double.parseDouble(next());}
        public void close(){
            try{in.close();}
            catch(IOException e){e.printStackTrace();}
        }
    }
}
