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? 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). For each test case output (n) integers, (i)-th of them representing the binary depth of the tree formed by the first (i) vertices. In the first test case, the fi