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"
#define rep(i, a) for (ll i = 0; i < (a); ++i)
#define REP(i, a, b) for (int i = (a); i < (b); ++i)
#define FORV(a, A) for (auto &a : A)

using namespace std;
using ll = long long;
using P = pair<int, int>;

const ll MOD = 1000000007;

void Main() {
  ll n;
  cin >> n;
  vector<ll> C(n);
  rep(i, n) cin >> C[i];
  sort(C.begin(), C.end());
  rep(i, n - 1)(C[i + 1] += C[i]) %= MOD;
  ll ans = C[n - 1];
  rep(i, n)(ans += C[i]) %= MOD;
  rep(i, 2 * n - 2)(ans *= 2) %= MOD;
  cout << ans << endl;
}

int main() {
  std::cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  std::cout << std::fixed << std::setprecision(15);
  Main();
  return 0;
}
