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;
using ll = long long;

const int maxn = 2e5 + 5;
ll pw[2*maxn];
const int mod = 1e9 + 7;

int main() {
    ios::sync_with_stdio(false);
//    cin.tie(0);

    pw[0] = 1;
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        pw[i] = (2*pw[i - 1]) % mod;
    }

    int c[maxn];
    for (int i = 1; i <= n; i++) {
        cin >> c[i];
    }
    sort(c + 1, c + n + 1);
    reverse(c + 1, c + n + 1);
    ll ans = (c[1]*pw[n - 1]) % mod;
    for (int j = 2; j <= n; j++) {
        ans += c[j] * ((pw[n - j] * ((pw[j - 2] * (j + 1)) % mod)) % mod);
        ans %= mod;
    }
    cout << (pw[n]*ans) % mod << '\n';

    return 0;
}