Problem Description


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


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


|Sodd−Seven|.


 



Sample Input


3 1 10 3 1 2 3 4 1 2 3 4


 



Sample Output


Hint

For the first sample, $A=\{10\}$, which contains one subset $\{10\}$ in which the number of elements is odd, and no subset in which the number of elements is even, therefore $S_{odd}=10,S_{even}=0,|S_{odd}-S_{even}|=10$.

For the second sample, $A=\{1,2,3\}$, which contains four subsets $\{1\},\{2\},\{3\},\{1,2,3\}$ in which the number of elements is odd, and three subsets $\{1,2\},\{2,3\},\{1,3\}$ in which the number of elements is even, therefore $S_{odd}=1+2+3+1=7,S_{even}=1+2+1=4,|S_{odd}-S_{even}|=3$.



可以根据题意暴力的走，但是实际上静下来想想就会发现其实答案就是最大值啊。

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<bitset>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 50;
int T, n, m, a[maxn];
LL c[maxn][maxn];

int main()
{
    c[0][0] = 1;
    for (int i = 1; i < maxn; i++)
    {
        c[i][0] = c[i][i] = 1;
        for (int j = 1; j < i; j++)
        {
            c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
        }
    }
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        for (int i = 0; i < n; i++) scanf("%d", &a[i]);
        sort(a, a + n);
        LL L = 0, R = 0;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; i + j < n; j++)
            {
                if (j & 1)
                {
                    R += c[n - i - 1][j] * a[i];
                }
                else
                {
                    L += c[n - i - 1][j] * a[i];
                }
            }
        }
        cout << abs(L - R) << endl;
    }
    return 0;
}
Close