Solve the programming task below in a Python markdown code block.
You are given a sequence $a_1, a_2, \ldots, a_n$. Each element of $a$ is $1$ or $2$.

Find out if an integer $k$ exists so that the following conditions are met.

$1 \leq k \leq n-1$, and

$a_1 \cdot a_2 \cdot \ldots \cdot a_k = a_{k+1} \cdot a_{k+2} \cdot \ldots \cdot a_n$.

If there exist multiple $k$ that satisfy the given condition, print the smallest.


-----Input-----

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). Description of the test cases follows.

The first line of each test case contains one integer $n$ ($2 \leq n \leq 1000$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 2$).


-----Output-----

For each test case, if there is no such $k$, print $-1$.

Otherwise, print the smallest possible $k$.


-----Examples-----

Input
3
6
2 2 1 2 1 2
3
1 2 1
4
1 1 1 1
Output
2
-1
1


-----Note-----

For the first test case, $k=2$ satisfies the condition since $a_1 \cdot a_2 = a_3 \cdot a_4 \cdot a_5 \cdot a_6 = 4$. $k=3$ also satisfies the given condition, but the smallest should be printed.

For the second test case, there is no $k$ that satisfies $a_1 \cdot a_2 \cdot \ldots \cdot a_k = a_{k+1} \cdot a_{k+2} \cdot \ldots \cdot a_n$

For the third test case, $k=1$, $2$, and $3$ satisfy the given condition, so the answer is $1$.