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.

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.

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

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.

题意：给一个长度为n的数组（数组元素没有重复，可以看成一个集合），求这个集合的子集中最大值和最小值的差，然后求和。

我们知道n个元素的集合的子集为2^n个，把数组从小到大排序后，任意的第i个元素和第j个元素(i<j)可以算出关系为2^(j-i-1)*(a[j]-a[i])，然后根据这个式子可以写出代码。

代码：

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

然而这个算法的时间复杂度为O(n^2)，很明显地会超时，不符合题目要求。

然后对a[i]这个元素进行观察和推算，可以推出差值为（2^(i-1)-2^(n-i)）*a[i]。

AC代码：

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 3e5 + 5;
typedef long long LL;
LL a[maxn], b[maxn];
#define mod 1000000007
int main()
{int n;while (cin >> n){int i;b[0] = 1;for (i = 1; i <= n; i++){cin >> a[i];b[i] = (2 * b[i - 1]) % mod;}sort(a + 1, a + 1 + n);LL ans = 0;for (i = 1; i <= n; i++)ans = (ans + (b[i - 1] - b[n - i])*a[i]) % mod;cout << ans << endl;}return 0;
}