codeforces809A Do 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 xi. 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, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that allxi are 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.

题解：

先将x从小到大排序

考虑一个集合的两个端点 如果端点固定下来那么这两个端点的贡献就是端点之间距离*端点之间点数的全集，

在计算答案时 考虑这样一种暴力：枚举端点之间距离 然后计算贡献 其实这样可以用前缀和优化掉 打表出2^300000%10000007的值 就可以实现O(n)

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=300000+10;
const long long MOD=1000000000+7;
int x[maxn];
typedef long long LL;
LL sum[maxn]={0};
inline LL quick_pow(LL x,LL y){
	if(y==0)
		return 1;
	if(y==1)
		return x;
	LL tmp=quick_pow(x,y/2)%MOD;
	tmp*=tmp;
	tmp%=MOD;
	if(y&1)
		tmp*=x;
	tmp%=MOD;
	return tmp;
}
int main(){
	//freopen("a.in","r",stdin);	
	//freopen("a.out","w",stdout);
	int n;
	scanf("%d",&n);
	LL ans=0;
	for(int i=1;i<=n;i++)
		scanf("%d",&x[i]);
	sort(x+1,x+n+1);
	for(int i=1;i<=n;i++)
		sum[i]=sum[i-1]+(LL)x[i];
	for(LL i=1;i<n;i++){
		LL u=quick_pow(2,i-1);
		u%=MOD;
		LL l=sum[n-i]%MOD;
		LL r=sum[n]%MOD-sum[i]%MOD;
		r%=MOD;
		if(r<0)
			r+=MOD;
		LL res=r%MOD-l%MOD;
		res%=MOD;
		if(res<0)
			res+=MOD;
		ans+=res*u%MOD;
		ans%=MOD;
		if(ans<0)
			ans+=MOD;
	}
	if(ans<0)
		ans+=MOD;
	printf("%I64d\n",ans%MOD);
return 0;
}