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 ll long long
#define fr(i,j,k) for(int i=j;i<k;i++)
#define f(n) fr(i,0,n)
#define f1(n) fr(i,1,n+1)
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
const int maxn = 1e5 + 5;
const int mod = 1e9+7;

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    ll four = 1;
    f(n - 1) {
        four *= 4;
        four %= mod;
    }
    vector<ll>c(n);
    for (auto &i : c) {
        cin >> i;
    }
    sort(all(c));
    reverse(all(c));
    ll mul = 0;
    f(n) {
        mul += c[i] * (i + 2);
        mul %= mod;
    }
    cout << four * mul % mod << '\n';
}