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 <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long ll;

const ll M = 1e9 + 7;

ll n;
ll a[200006];
ll ans;
ll now = 1;
ll x = 1;

int main()
{
    scanf("%lld", &n);
    for(ll i = 0; i < n; i++)
    {
        scanf("%lld", &a[i]);
        if(i >= 2) x *= 2, x %= M;
    }
    if(n == 1)
    {
        printf("%lld", a[0] * 2 % M);
        return 0;
    }
    sort(a, a + n, greater<ll>());
    now = x;
    for(ll i = 0; i < n; i++)
    {
        now += x;
        now %= M;
        ans += now * a[i] % M;
        ans %= M;
    }
    printf("%lld", ans * x % M * 4 % M);
    return 0;
}
