Codeforces Round #415 (Div. 2) C. Do you want a date? 數學

題目：

C. 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 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 all xi 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 0 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.

這個題目在紙上畫一畫，思路還是很好找的。比如{1,2,3,4}，它是讓求每個集合的極差之和，我們只考慮每個集合中的最大和最小就好了。我們固定這個集合中的最大值和最小值，(與之相應的，在這之前要排序)比如說，最大是4，最小是1，那麼有多少個集合滿足這樣的情況呢?答案是4。爲什麼？你想啊，在1,4中間有兩個數，這兩個數可以選或者不選2*2=4。那麼思路就很清晰了，先sort，固定起點i,終點j。(i<j)再枚舉，就好了。複雜度是O(n^2+nlogn)絕對過不了的。

好吧，我們先把等式寫出來。res+=(a[i]-a[j])*2^(i-j-1).我們可以試着手動化簡一下，那麼局勢就更加明朗了。res+=(a[n]-a[1])*2^0+(a[n]+a[n-1]-a[1]-a[2])*2+(sigma a[n-2]..a[n] - sigama a[1]...a[3])*2^3.   我們可以很容易的想到前綴數組。  

最後，處理這個2的n次冪也是很重要的一點，千萬不能中二病，自己去寫一個算power(2,n)那麼複雜度也就達到了O(n^2)，正確的寫法是在循環中更新。同時不要忘了mod，我就是因爲這個WA了兩發。

code：

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MAXN=3e5+5;
const int mod=1e9+7;
LL sum[MAXN],a[MAXN];
int main(){
    int n;scanf("%d",&n);
    for(int i=1;i<=n;++i)
        scanf("%I64d",&a[i]);
    sort(a+1,a+n+1);
    for(int i=1;i<=n;++i)
        sum[i]=sum[i-1]+a[i];
    long long res=0;
    LL weight=1;
    for(int i=1;i<=n;++i){
        res=(res+( (sum[n]-sum[n-i]) -sum[i])%mod*weight%mod)%mod;
        weight*=2;
        weight%=mod;
    }
    printf("%I64d\n",res);
    return 0;
}