CodeForces809ADo you want a date?

Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.

Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to n. So the i-th hacked computer is located at the point x_i. Moreover the coordinates of all computers are distinct.

Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.

Leha should calculate a sum of F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression . Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, . Since the required sum can be quite large Noora asks to find it modulo 10⁹ + 7.

Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.

Input

The first line contains one integer n (1 ≤ n ≤ 3·10⁵) denoting the number of hacked computers.

The second line contains n integers x₁, x₂, ..., x_n (1 ≤ x_i ≤ 10⁹) denoting the coordinates of hacked computers. It is guaranteed that all x_iare distinct.

Output

Print a single integer — the required sum modulo 10⁹ + 7.

Examples

input

2
4 7

output

3

input

3
4 3 1

output

9

Note

There are three non-empty subsets in the first sample test:,  and . The first and the second subset increase the sum by 0and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.

There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: , , , . In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.

给你一个n个数的数组，在这些数组成的任意非空集合中，用每个集合中的最大值减最小值，求这些值的总和，另外还要对1e9+7求模。

首先想到的是一个非常简单的规律，在对数组从小到大排序后，任意第i个数和第j个数(i<j)的关系为2^(j-i-1)*(a[j]-a[i])，然后就得出可以满足答案逻辑的一串代码

for(i=1;i<=n;i++)
    {
        for(j=i+1;j<=n;j++)
        {
            k=j-i-1;
            ans+=b[k]*(a[j]-a[i])%mod;
            ans%=mod;
        }
    }注：数组b为2^(n-1)%mod

这段代码在逻辑上可行，但时间复杂度为o(n^2)，不能满足题目要求，因此要对规律进行进一步的优化。

观察第i个数与其余所有数之间的关系，并且只统计第i个数的加减情况，那么a[i]的情况为2^(i-1)-2^(n-i)次

完整代码如下

#include<bits/stdc++.h>
using namespace std;
#define MAX 300000
#define mod 1000000007
long long a[MAX+5],b[MAX+5],ans=0;

int main()
{
 int n,i;
 scanf("%d",&n);
 b[0]=1;
 for(i=1;i<=n;i++)
 {
  scanf("%d",&a[i]);
  b[i]=(2*b[i-1])%mod;
 }
 sort(a+1,a+n+1);
 for(i=1;i<=n;i++)
 {
  ans+=(b[i-1]-b[n-i])*a[i]%mod;
  ans%=mod;
 }
 cout<<ans<<endl;
 }

 另外，对于核心思路，还有另一种解法，思路略有不同，但优势在于不需要建立一个幂次方的数组。

    for(i=2;i<=n;i++)
    {
        ans=(ans+a[i]*(f-1)-sum)%mod;
        sum=(2*sum+a[i])%mod;
        f=(f*2)%mod;
    }