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;

ll MOD = 1e9 + 7;

ll calc(ll x) {
  if (x == 0) return 1;
  if (x == 1) return 2;
  ll ans = calc(x / 2);
  ans = (ans * ans) % MOD;
  if (x % 2 == 1) ans = (ans * 2) % MOD;
  return ans;
}

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

  ll N; cin >> N;
  vector<ll> C(N); for (ll i = 0; i < N; i++) cin >> C[i];
  sort(C.begin(), C.end());

  ll ans = 0;
  for (int i = 0; i < N; i++) {
    ll M = N - i;
    ll temp = ((N - i - 1) * calc(2 * N - 2)) % MOD;
    temp = (temp + calc(2 * N - 1)) % MOD;
    temp = (temp * C[i]) % MOD;
    ans = (ans + temp) % MOD;
  }
  cout << ans << '\n';
  return 0;
}