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;
const int maxn = 2e5 + 10, mod = 1e9 + 7;
int n, ans, t, s;
int c[maxn];
ll fpm(ll a, ll b) {
	ll ret = 1ll;
	for (; b; b >>= 1, a = a * a % mod)
		if (b & 1)ret = ret * a % mod;
	return ret;
}
void solve() {
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; ++i) cin >> c[i];
	sort(c + 1, c + n + 1, greater<int>());
	t = fpm(2, 2 * n - 2);
	for (int i = 1; i <= n; ++i)
		(ans += 1ll * (i + 1) * c[i] % mod) %= mod;
	ans = 1ll * ans * t % mod;
	cout << ans << endl;
}
int main() {
	solve();
	return 0;
}