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 ll long long
using namespace std;
const ll maxn = 2e5 + 5 , mod = 1e9 + 7;
int n;
ll c[maxn];
ll fpow(ll x , ll k) {
	ll ans = 1 , base = x;
	while(k) {
		if(k & 1) ans = ans * base % mod;
		base = base * base % mod;
		k >>= 1;
	}
	return ans;
}
inline bool cmp(ll a , ll b){ return a > b; }
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cin >> n;
	for(register int i = 1; i <= n; ++i) cin >> c[i];
	sort(c + 1 , c + n + 1 , cmp);
	ll ans = 0;
	for(register int i = 1; i <= n; ++i) 
		ans = (ans + c[i] * (i + 1) % mod) % mod;
	ans = ans * fpow(2 , 2 * n - 2) % mod;
	cout << ans << endl;
	return 0;
} 