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;

#define watch(x) cerr << (#x) << ": " << (x) << endl
#define int long long

const int N = 200009, mod = 1000000007;
int n, c[N];

int pow(int a, int b) {
    if(b == 0) return 1;
    if(b%2) return a*pow(a, b-1)%mod;
    return pow(a*a%mod, b/2)%mod;
}

int32_t main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n;
    for(int i = 0; i < n; ++i) {
        cin >> c[i];
    }

    sort(c, c+n, greater<int>());

    int ans = 0;
    for(int i = 0; i < n; ++i) {
        ans = (ans + c[i]*(i+2))%mod;
    }

    ans = (ans*pow(4LL, n-1))%mod;

    cout << ans << endl;
    
    return 0;
}
