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<stdio.h>
#include<algorithm>
#define M 1000000007
using namespace std;

long long pow[200001];
long long C[200000];

int main() {
    int n, i;
    scanf("%d", &n);

    for(i = 0; i < n; i++) scanf("%lld", &C[i]);
    sort(C, C + n);
    reverse(C, C + n);

    pow[0] = 1;
    for(i = 1; i <= n; i++) pow[i] = (pow[i - 1] * 2) % M;

    long long ans = 0;
    for(i = 0; i < n; i++) {
        long long res = pow[i];
        if (i > 0) res = (res + pow[i - 1] * i) % M;
        res = (res * C[i]) % M;
        res = (res * pow[n - i - 1]) % M;
        ans = (ans + res) % M;
    }

    printf("%lld\n", (ans * pow[n]) % M);
    return 0;
}