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,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int mod = 1000000007;

int main() {
    int n;
    cin >> n;
    vector<int> C(n);
    rep(i, n) cin >> C[i];
    sort(C.begin(), C.end(), greater<int>());
    vector<ll> twoexp(n);
    twoexp[0] = 1;
    rep(i, n-1) twoexp[i+1] = (twoexp[i]*2)%mod;
    ll ans = 0;
    rep(i, n) {
      ll tmp = twoexp[i];
      if (i > 0) tmp += twoexp[i-1] * i;
      tmp %= mod;
      tmp *= C[i];
      tmp %= mod;
      tmp *= twoexp[n-i-1];
      tmp %= mod;
      ans += tmp;
      ans %= mod;
    }
    ans *= twoexp[n-1] * 2;
    ans %= mod;
    cout << ans << endl;
}