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<algorithm>
#include<cstdio>
using namespace std;

const int MAX_N = 2e5, A = 1e9 + 7;
int n;
int c[MAX_N];

int po(int x, int y) {
    int ans = 1;
    while (y > 0) {
        if (y & 1) ans = static_cast<long>(ans) * x % A;
        x = static_cast<long>(x) * x % A;
        y >>= 1;
    }
    return ans;
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &c[i]);
    sort(c, c + n);
    
    int ans = 0;
    for (int i = 0; i < n; i++) {
        ans = (ans + (po(2, 2 * (n - i - 1))
            + static_cast<long>(po(2, 2 * (n - i - 1) - 1))
            * (n - i - 1) % A) % A * po(2, 2 * i + 1) % A * c[i] % A) % A;
    }
    printf("%d\n", ans);
    return 0;
}
