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
#pragma GCC optimize("Ofast", "unroll-loops")

#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ld long double

const ll mod = 1000000007LL;

ll modpow(ll a, ll b) {
	if (b == 0) return 1;
	ll t = modpow(a, b / 2);
	if (b & 1)
		return t * t % mod * a % mod;
	return t * t % mod;
}

signed main() {
	ll n; cin >> n;
	vector<ll> c(n);
	for (int i = 0; i < n; ++i) cin >> c[i];
	sort(c.begin(), c.end());

	ll ans = 0;
	for (int i = 0; i < n; ++i) {
		ll t = c[i] * (-i + n + 1) % mod;
		ans += t; ans %= mod;
	}
	ans *= modpow(2, 2 * n - 2LL);
	ans %= mod;
	cout << ans << endl;
	
	return 0;
}