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;

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    int n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    sort(a.begin(), a.end());
    long long MOD = 1e9 + 7;
    vector<long long> pw2(2*n+1, 1);
    for (int i = 1; i < pw2.size(); i++) {
        pw2[i] = (pw2[i-1] * 2) % MOD;
    }
    long long cur = 0;
    for (int i = n-1; i >= 0; i--) {
        int k = n-1-i;
        cur = (a[i]*(2*(k == 0 ? 1ll : pw2[k-1+k]*(k+2)) % MOD)%MOD + 4*cur)%MOD;
    }
    cout << cur << endl;
}

