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 <bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
int c[200005], p2[200005], p4[200005];
int func(int l){
    if(l == 0) return 0;
    return l*1ll*p2[l-1]%mod;
}
int main(){
    int n;
    scanf("%d", &n);
    p2[0] = p4[0] = 1;
    for(int i=1;i<=n;i++){
        scanf("%d", &c[i]);
        p2[i] = p2[i-1]*2%mod;
        p4[i] = p4[i-1]*1ll*4%mod;
    }   
    sort(c+1, c+1+n);
    int ans = 0;
    for(int i=1;i<=n;i++){
        int l = i-1, r = n-i;
        int res = p4[l]*1ll*p2[r+1]%mod*(func(r) + p2[r])%mod*c[i]%mod;
        ans = (ans + res)%mod;
    }
    printf("%d\n", ans);
}