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;

typedef long long int ll;

const int mod = 1e9 + 7;

ll pwr (ll a, ll b) {
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}

int main () {
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  int n;
  cin >> n;
  vector < int > c (n + 1);
  for (int i = 1; i <= n; i++) cin >> c[i];
  sort (c.begin() + 1, c.end(), greater < int > ());

  ll ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += (ll)c[i] * (i + 1);
    ans %= mod;
  }
  ans = (ans * pwr (4, n - 1)) % mod;

  cout << ans << '\n';
}
