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 ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=5e15;

ll mpow(ll x, ll n) {
  ll ans=1;
  while(n>0) {
    if(n&1) {
      ans=ans*x%mod;
    }
    x=x*x%mod;
    n>>=1;
  }
  return ans;
}
 
int main() {
  ll n;
  cin >> n;
  vec c(n);
  for(ll i=0;i<n;i++) cin >> c[i];
  sort(c.rbegin(),c.rend());
  ll ans=0;
  for(ll i=0;i<n;i++) {
    (ans+=c[i]*(i+2)%mod)%=mod;
  }
  (ans*=mpow(4,n-1))%=mod;
  cout << ans << endl;
}