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 <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;
const int SIZEN = 200010;
LL pw[SIZEN];
LL C[SIZEN];
int N;
int main(){
    scanf("%d",&N);
    pw[0] = 1;
    for(int i = 1;i <= N;i++)pw[i] = pw[i-1] * 2 % mod;
    for(int i = 1;i <= N;i++)scanf("%lld",&C[i]);
    sort(C+1,C+1+N);
    LL ans = 0;
    for(int i = 1;i <= N;i++){
        ans = (ans + C[i] * pw[i-1] % mod * pw[N]%mod * ( (N-i)  * pw[max(N-i-1,0)]%mod + pw[N-i] ) % mod ) % mod;
    }
    printf("%lld\n",ans);
    return 0;
}