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 i64 = int64_t;

constexpr i64 MOD = 1e9 + 7;

i64 repeated_squaring(i64 x, i64 n) {
    i64 res = 1;
    while (n > 0) {
        if (n & 1)
            res = res * x % MOD;
        x = x * x % MOD;
        n >>= 1;
    }
    return res;
}

int main(){
    i64 n;
    cin >> n;
    vector<i64> v;
    for(int i=0;i<n;++i){
        i64 c;
        cin >> c;
        v.push_back(c);
    }
    sort(v.rbegin(), v.rend());
    i64 ans = 0;
    for(i64 k=1;k<=n;++k){
        ans += (v[k-1] * (k+1)) % MOD;
        ans %= MOD;
    }

    ans *= repeated_squaring(4, n-1);
    ans %= MOD;
    cout << ans << endl;

    return 0;
}