CF1933A.Turtle Puzzle: Rearrange and Negate

传统题 时间 2000 ms 内存 256 MiB 3 尝试 1 已通过 1 标签

Turtle Puzzle: Rearrange and Negate

You are given an array aa of nn integers. You must perform the following two operations on the array (the first, then the second):

  1. Arbitrarily rearrange the elements of the array or leave the order of its elements unchanged.
  2. Choose at most one contiguous segment of elements and replace the signs of all elements in this segment with their opposites. Formally, you can choose a pair of indices l,rl, r such that 1lrn1 \le l \le r \le n and assign ai=aia_i = -a_i for all lirl \le i \le r (negate elements). Note that you may choose not to select a pair of indices and leave all the signs of the elements unchanged.

What is the maximum sum of the array elements after performing these two operations (the first, then the second)?

Input

The first line of the input contains a single integer tt (1t10001 \le t \le 1000) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case contains a single integer nn (1n501 \le n \le 50) — the number of elements in array aa.

The second line of each test case contains nn integers a1,a2,,ana_1, a_2, \ldots, a_n (100ai100-100 \le a_i \le 100) — elements of the array.

Output

For each test case, output the maximum sum of the array elements after sequentially performing the two given operations.

Note

In the first test case, you can first rearrange the array to get [3,2,3][3,-2,-3] (operation 1), then choose l=2,r=3l = 2, r = 3 and get the sum 3+((2)+(3))=83 + -((-2) + (-3)) = 8 (operation 2).

In the second test case, you can do nothing in both operations and get the sum 00.

In the third test case, you can do nothing in both operations and get the sum 0+1=10 + 1 = 1.

In the fourth test case, you can first leave the order unchanged (operation 1), then choose l=1,r=1l = 1, r = 1 and get the sum (99)=99-(-99) = 99 (operation 2).

In the fifth test case, you can first leave the order unchanged (operation 1), then choose l=2,r=3l = 2, r = 3 and get the sum 10+((2)+(3))+7=2210 + -((-2) + (-3)) + 7 = 22 (operation 2).

In the sixth test case, you can first leave the order unchanged (operation 1), then choose l=1,r=5l = 1, r = 5 and get the sum ((1)+(2)+(3)+(4)+(5))=15-((-1)+(-2)+(-3)+(-4)+(-5))=15 (operation 2).

Samples

8
3
-2 3 -3
1
0
2
0 1
1
-99
4
10 -2 -3 7
5
-1 -2 -3 -4 -5
6
-41 22 -69 73 -15 -50
12
1 2 3 4 5 6 7 8 9 10 11 12
8
0
1
99
22
15
270
78

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