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;
#define ll long long

ll mod = 1e9+7;

int main() {
    int N;
    cin >> N;
    vector<ll> C(N);
    for (int i = 0; i < N; i++) cin >> C[i];
    sort(C.begin(), C.end());
    vector<ll> pow(N+1, 1);
    for (int i = 1; i <= N; i++) pow[i] = pow[i-1]*2%mod;
    vector<ll> sum(N+1, 0);
    sum[0] = C[0];
    for (int i = 1; i < N; i++){
        sum[i] = (sum[i-1]+C[i])%mod;
    }
    vector<ll> dp(N, 0);
    dp[0] = C[0]*2%mod;
    for (int i = 1; i < N; i++) {
        dp[i] = dp[i-1]*4;
        dp[i] += C[i]*2*pow[i]%mod*pow[i]%mod;
        dp[i] += sum[i-1]*4*pow[i-1]%mod*pow[i-1]%mod;
        dp[i] %= mod;
    }
    cout << dp[N-1] << endl;
    return 0;
}
