zxa and set

题目连接：

Description

zxa has a set A={a1,a2,⋯,an}, which has n elements and obviously (2n−1) non-empty subsets.

For each subset B={b1,b2,⋯,bm}(1≤m≤n) of A, which has m elements, zxa defined its value as min(b1,b2,⋯,bm).

zxa is interested to know, assuming that Sodd represents the sum of the values of the non-empty sets, in which each set B is a subset of A and the number of elements in B is odd, and Seven represents the sum of the values of the non-empty sets, in which each set B is a subset of A and the number of elements in B is even, then what is the value of |Sodd−Seven|, can you help him?

Input

The first line contains an positive integer T, represents there are T test cases.

For each test case:

The first line contains an positive integer n, represents the number of the set A is n.

The second line contains n distinct positive integers, repersent the elements a1,a2,⋯,an.

There is a blank between each integer with no other extra space in one line.

1≤T≤100,1≤n≤30,1≤ai≤109

Output

For each test case, output in one line a non-negative integer, repersent the value of |Sodd−Seven|.

Sample Input

3

1

10

3

1 2 3

4

1 2 3 4

Sample Output

10

3

4

Hint

题意

zxa有一个集合\(A=\{a_1,a_2,\cdots,a_n\}\)，\(n\)表示集合\(A\)的元素个数，这个集合明显有\((2^n-1)\)个非空子集合。

对于每个属于\(A\)的子集合\(B=\{b_1,b_2,\cdots,b_m\}(1\leq m\leq n)\)，\(m\)表示集合\(B\)的元素个数，zxa定义它的价值是\(\min(b_1,b_2,\cdots,b_m)\)。

zxa很好奇，如果令\(S_{odd}\)表示集合\(A\)的所有含奇数个元素的非空子集合的价值之和，\(S_{even}\)表示集合\(A\)的所有含偶数个元素的非空子集合的价值之和，那么\(|S_{odd}-S_{even}|\)是多少，你能帮助他吗？

题解：

不管是数学方法，或者瞎爆搜，可以猜到答案就是最大值

直接输出就好了

代码

#include
    
     using namespace std;void solve(){    int n;scanf("%d",&n);    int ans=0,x;    for(int i=1;i<=n;i++)    {        scanf("%d",&x);        ans=max(ans,x);    }    printf("%d\n",ans);}int main(){    int t;    scanf("%d",&t);    while(t--)solve();    return 0;}