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 <stdio.h>
#include <algorithm>
#include <functional>
#define mod 1000000007
using namespace std;
typedef long long ll;

ll mpow(ll a, ll b) {
  ll ans = 1;
  while(b) {
    if(b & 1) ans *= a, ans %= mod;
    a *= a, a %= mod, b >>= 1;
  }
  return ans;
}

int main(void) {
  ll i, j, k, n, ans = 0;
  scanf("%lld", &n);
  ll c[n];
  for(i = 0; i < n; ++i) scanf("%lld", &c[i]);
  sort(c, c + n, greater<ll>());
  for(i = 0; i < n; ++i) {
    k = i * (i ? mpow(2, i - 1) : 0) + mpow(2, i), k %= mod;
    k *= c[i], k %= mod;
    k *= mpow(2, 2 * n - i - 1), k %= mod;
    ans += k, ans %= mod;
  }
  printf("%lld", ans);
  return 0;
}