CF1300A.Non-zero

Guy-Manuel and Thomas have an array $a$ of $n$ integers [$a_1, a_2, \dots, a_n$]. In one step they can add $1$ to any element of the array. Formally, in one step they can choose any integer index $i$ ($1 \le i \le n$) and do $a_i := a_i + 1$.

If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.

What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make $a_1 + a_2 +$ $\dots$ $+ a_n \ne 0$ and $a_1 \cdot a_2 \cdot$ $\dots$ $\cdot a_n \ne 0$.

盖伊-曼努埃尔和托马斯有一个包含 $n$ 个整数的数组 $a$：$[a_1, a_2, \dots, a_n]$。每一步中，他们可以将数组中任意一个元素加 $1$。形式化地说，每一步中他们可以选择任意下标 $i$（满足 $1 \le i \le n$），并执行操作 $a_i := a_i + 1$。

如果数组所有元素的和或积等于零，盖伊-曼努埃尔和托马斯并不介意再多执行一次该操作。

那么，他们至少需要多少步，才能使得数组所有元素的和与积均不为零？即：求最小操作步数，使得

$$a_1 + a_2 + \dots + a_n \ne 0 \quad \text{且} \quad a_1 \cdot a_2 \cdot \dots \cdot a_n \ne 0。$$

Each test contains multiple test cases.

The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 100$) — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($-100 \le a_i \le 100$) — elements of the array .

每个测试包含多个测试用例。

第一行包含测试用例的数量 $t$（$1 \le t \le 10^3$）。随后是测试用例的描述。

每个测试用例的第一行包含一个整数 $n$（$1 \le n \le 100$）—— 数组的大小。

每个测试用例的第二行包含 $n$ 个整数 $a_1, a_2, \dots, a_n$（$-100 \le a_i \le 100$）—— 数组的元素。

For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.

对于每个测试用例，输出使数组中所有元素的和与积均不为零所需的最少操作步数。

In the first test case, the sum is $0$. If we add $1$ to the first element, the array will be $[3,-1,-1]$, the sum will be equal to $1$ and the product will be equal to $3$.

In the second test case, both product and sum are $0$. If we add $1$ to the second and the third element, the array will be $[-1,1,1,1]$, the sum will be equal to $2$ and the product will be equal to $-1$. It can be shown that fewer steps can't be enough.

In the third test case, both sum and product are non-zero, we don't need to do anything.

In the fourth test case, after adding $1$ twice to the first element the array will be $[2,-2,1]$, the sum will be $1$ and the product will be $-4$.

在第一个测试用例中，和为 $0$。如果我们将 $1$ 加到第一个元素上，则数组变为 $[3,-1,-1]$，此时和为 $1$，积为 $3$。

在第二个测试用例中，积与和均为 $0$。如果我们将 $1$ 分别加到第二个和第三个元素上，则数组变为 $[-1,1,1,1]$，此时和为 $2$，积为 $-1$。可以证明，无法用少于该次数的操作达成目标。

在第三个测试用例中，和与积均非零，因此无需进行任何操作。

在第四个测试用例中，将 $1$ 两次加到第一个元素后，数组变为 $[2,-2,1]$，此时和为 $1$，积为 $-4$。