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>
typedef long long lint;
using namespace std;

const lint mod = 1e+9 + 7;
lint n;
vector<lint> c;

// a^n mod pを計算　O(log n)
long long pow_mod(long long a, long long n, long long p) {
    if (n == 0) return 1;
    if (n & 1) return pow_mod(a, n - 1, p) * a % p;
    long long temp = pow_mod(a, n / 2, p);
    return temp * temp % p;
}

int main() {
    cin >> n;
    c.resize(n);
    for (int i = 0; i < n; i++) {
        cin >> c[i];
    }
    sort(c.begin(), c.end());
    lint pow = pow_mod(4, n - 1, mod);
    lint ans = 0;
    for (int i = 0; i < n; i++) {
        lint add = pow * (n - i + 1) % mod;
        add *= c[i];
        ans = (ans + add) % mod;
    }
    cout << ans << endl;
}