Codeforces Round #415 (Div. 1) A Do you want a date?

A. Do you want a date?

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ton 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 from1 to n. So thei-th hacked computer is located at the pointx_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 alla, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denoteA the set of all integers from 1 to n. Noora asks the hacker to find value of the expression. HereF(a) is calculated as the maximum among the distances between all pairs of computers from the seta. 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_i are distinct.

Output

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

Examples

Input

24 7

Output

3

Input

34 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 by0 and 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.

题目很长，然而信息很少，六级没过的还是回去练练吧。

题目大意：给一个数组，问你所有（非空子集中最大之减去最小值）之和。

一开始想线段树？不行。动归？时间复杂度太高了但是确实像是动归题，没什么思路。

然后开始分析题意：{1,2,3,4,5}这个集合里，最后的sum是(2-1)*1+(3-1)*2+(4-1)*4+(5-1)*8+(3-2)*1+(4-2)*2+(5-2)*4+(4-3)+(5-3)*2+5-4

这里我们已经可以看出规律了

更进一步，把所有的最大值最小值都不看，只看每个数字加了多少次，

1:-15=1-16

2:-6,=2-8

3:0=4-4

4:6=8-2

5:15=16-1

所以从小到大顺序的数列第i个数字加的次数为1<<(i-1)-1<<(n-i)，这个结果乘以这个数字就是贡献度了。

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;typedef long long ll;const int mod = 1e9+7;ll a[300086];ll s[300086];int main(){    int n;    cin>>n;    s[0]=1;    for(int i=1;i<=n;i++){        cin>>a[i];        s[i]=(s[i-1]*2)%mod;    }    sort(a+1,a+n+1);    ll sum = 0;    for(int i=n;i>0;i--){        sum=sum+((s[i-1]-s[n-i])%mod)*a[i];        sum%=mod;    }    while(sum<0)        sum+=mod;    cout<<sum<<endl;    return 0;}