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.*;

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };

	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int n = fs.nextInt();
		int c[] = fs.nextIntArray(n);
		Arrays.sort(c);
		
		int four[] = new int[n];
		four[0]=1;
		for(int i=0;i<n-1;i++) {
			four[i+1] = (int)mul(four[i],4); 
		}
		long ans = 0;
		for(int i=0;i<n;i++) {
			long d = mul(four[n-1],n-i+1);
			long tmp = mul(c[i],d);
			ans = plus(ans, tmp);
		}
		System.out.println(ans);
		
	}
	
	//reverseOrderSort : O(N)
	static void rsort(int[] ar){
		Arrays.sort(ar);
		int len = ar.length;
		for(int i=0;i<len/2;i++) {
			int tmp = ar[i];
			ar[i] = ar[len-1-i];
			ar[len-1-i] = tmp;
		}
	}
	static void rsort(long[] ar){
		Arrays.sort(ar);
		int len = ar.length;
		for(int i=0;i<len/2;i++) {
			long tmp = ar[i];
			ar[i] = ar[len-1-i];
			ar[len-1-i] = tmp;
		}
	}
	
	//MOD culculations
	static long plus(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x + y) % mod;
		return res;
	}

	static long suv(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x - y + mod) % mod;
		return res;
	}

	static long mul(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * y % mod;
		return res;
	}

	static long div(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * modpow(y, mod - 2) % mod;
		return res;
	}

	//modの元での高速pow()
	static long modpow(long a, long p) {
		if (p == 0)
			return 1;
		if (p % 2 == 0) {
			long halfP = p / 2;
			long root = modpow(a, halfP);
			return root * root % mod;
		} else
			return a * modpow(a, p - 1) % mod;
	}

}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}