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
//In The Name of Allah
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 2e5 + 10;
const ll mod = 1e9 + 7;

ll c[N], pw[N];

ll f(ll x){
    if (x == 0)
        return 1;
    return (x + 2) * pw[x - 1] % mod;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    ll n;
    cin >> n;
    for (ll i = 1; i <= n; i ++)
        cin >> c[i];
    sort(c + 1, c + n + 1);
    pw[0] = 1;
    for (ll i = 1; i < N; i ++)
        pw[i] = pw[i - 1] * 2 % mod;
    ll ans = 0;
    for (ll i = 1; i <= n; i ++)
        ans = (ans + f(n - i) * pw[i - 1] % mod * c[i] % mod) % mod;
    ans = ans * pw[n] % mod;
    cout << ans;
    return 0;
}
