B. The 67th 6-7 Integer Problem

Felven2026-04-21 13:33

time limit per test

1 second

memory limit per test

256 megabytes

So, Macaque has passed his first challenge (and is not acknowledging your help whatsoever). After all, it was only given to him so he could engage in his greatest pleasure --- crunching on desiccated freezedried hamburgers and yelling 'trivial' at the screen. However, he has another, much more important task ahead of him, and he has once again enlisted you to help him.

You are given 7 integers a1,a2,...,a7.

You must negate 6 out of the 7 integers (that is, multiply them by −1). Over all possible ways to negate 6 out of the 7 integers, find the maximum possible sum of a.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤6767). The description of the test cases follows.

The first and only line of each test case contains 7 space-separated integers a1,a2,...,a7 (−67≤ai≤67).

Output

For each test case, output the maximum sum of a after negating 6 out of the 7 integers, on a new line.

Example

Input

Copy

复制代码

41 41 41 41 41 41 41

6 9 4 20 6 7 67

1 2 3 4 5 6 7

6 7 6 7 6 7 6

Output

Copy

复制代码

-205

-14

-31

Note

In the first test case, no matter which 6 integers we pick to negate, the maximum sum is −41−41−41−41−41−41+41=−205.

In the second test case, we can negate all integers except a7 to obtain a sum of −6−9−4−20−6−7+67=15.

解题说明：此题其实就是找出最大的那个数，然后其他数都变成相反数，最后求和。

cpp 复制代码

#include <stdio.h>

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int a[7];
        int sum = 0, mx = -100;
        for (int i = 0; i < 7; i++)
        {
            scanf("%d", &a[i]);
            sum += a[i];
            if (a[i] > mx)
            {
                mx = a[i];
            }
        }
        int ans = 2 * mx - sum;
        printf("%d\n", ans);
    }
    return 0;
}