Iris likes full binary trees.

Let's define the depth of a rooted tree as the maximum number of vertices on the simple paths from some vertex to the root. A full binary tree of depth $$$d$$$ is a binary tree of depth $$$d$$$ with exactly $$$2^d - 1$$$ vertices.

Iris calls a tree a $$$d$$$-binary tree if some vertices and edges can be added to it to make it a full binary tree of depth $$$d$$$. Note that any vertex can be chosen as the root of a full binary tree.

Since performing operations on large trees is difficult, she defines the binary depth of a tree as the minimum $$$d$$$ satisfying that the tree is $$$d$$$-binary. Specifically, if there is no integer $$$d \ge 1$$$ such that the tree is $$$d$$$-binary, the binary depth of the tree is $$$-1$$$.

Iris now has a tree consisting of only vertex $$$1$$$. She wants to add $$$n - 1$$$ more vertices to form a larger tree. She will add the vertices one by one. When she adds vertex $$$i$$$ ($$$2 \leq i \leq n$$$), she'll give you an integer $$$p_i$$$ ($$$1 \leq p_i < i$$$), and add a new edge connecting vertices $$$i$$$ and $$$p_i$$$.

Iris wants to ask you the binary depth of the tree formed by the first $$$i$$$ vertices for each $$$1 \le i \le n$$$. Can you tell her the answer?

Input format:
Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \leq n \leq 5 \cdot 10^5$$$) — the final size of the tree.

The second line of each test case contains $$$n - 1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \leq p_i < i$$$) — descriptions of all edges of the tree.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.
Output format:
For each test case output $$$n$$$ integers, $$$i$$$-th of them representing the binary depth of the tree formed by the first $$$i$$$ vertices.
Example Input 0:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23

Example Output 0:
1 2 2 
1 2 2 3 3 4 
1 2 2 3 3 4 4 
1 2 2 3 3 3 4 4 5 5 
1 2 2 3 3 4 4 4 -1 -1 
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8

Notes:
In the first test case, the final tree is shown below:

- The tree consisting of the vertex $$$1$$$ has the binary depth $$$1$$$ (the tree itself is a full binary tree of depth $$$1$$$).
- The tree consisting of the vertices $$$1$$$ and $$$2$$$ has the binary depth $$$2$$$ (we can add the vertex $$$3$$$ to make it a full binary tree of depth $$$2$$$).
- The tree consisting of the vertices $$$1$$$, $$$2$$$ and $$$3$$$ has the binary depth $$$2$$$ (the tree itself is a full binary tree of depth $$$2$$$).

In the second test case, the formed full binary tree after adding some vertices to the tree consisting of $$$n$$$ vertices is shown below (bolded vertices are added):

The depth of the formed full binary tree is $$$4$$$.

In the fifth test case, the final tree is shown below:

It can be proved that Iris can't form any full binary tree by adding vertices and edges, so the binary depth is $$$-1$$$.

Example input:
7
3
1 1
6
1 2 3 4 5
7
1 1 3 2 5 1
10
1 1 2 1 4 2 4 5 8
10
1 1 3 1 3 2 2 2 6
20
1 1 2 2 4 4 5 5 7 6 8 6 11 14 11 8 13 13 12
25
1 1 3 3 1 5 4 4 6 8 11 12 8 7 11 13 7 13 15 6 19 14 10 23


Example output:
1 2 2 
1 2 2 3 3 4 
1 2 2 3 3 4 4 
1 2 2 3 3 3 4 4 5 5 
1 2 2 3 3 4 4 4 -1 -1 
1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 6 6 7 
1 2 2 3 3 4 4 4 4 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8