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;
const int mod = 1e9+7;
long long ans;
long long modPow(long long a, int p){
    if (p == 0) return 1;
    if (p%2 == 0) {
        long long half = modPow(a, p/2);
        return half * half % mod;
    }
    else return a * modPow(a, p-1) % mod;
}
int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vector<int> C(N);
    for (auto & c : C) cin >> c;
    sort(C.rbegin(), C.rend());
    for (int i = 0; i < N; ++i) {
        ans += (long long)C[i]*(i+2);
        ans %= mod;
    }
    ans *= modPow(4, N-1);
    ans %= mod;
    printf("%lld\n", ans);
    return 0;
}
