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 ll = long long;
static const ll MOD = 1e9 + 7;

// 繰り返し自乗法(x^nをmで割った余り) O(logn)
ll power(ll x, ll n, ll m) {
  ll res = 1;
  if (n > 0) {
    res = power(x, n / 2, m);
    if(n % 2 == 0) res = (res * res) % m;
    else res = (((res * res) % m) * x) % m;
  }
  return res;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n;
  cin >> n;
  
  vector<ll> c(n);
  for (int i = 0; i < n; i++) cin >> c.at(i);
  
  sort(c.rbegin(), c.rend());
  
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    ans += (((i + 2) * c.at(i)) % MOD * power(4, n - 1, MOD)) % MOD;
    ans %= MOD;
  }
  
  cout << ans << '\n';
}