E. You
dfs and similar; dp; math; number theory
You are given a tree with \(n\) nodes. As a reminder, a tree is a connected undirected graph without cycles.Let \(a_1, a_2, \ldots, a_n\) be a sequence of integers. Perform the following operation exactly \(n\) times: Select an unerased node \(u\). Assign \(a_u :=\) number of unerased nodes adjacent to \(u\). Then, erase the node \(u\) along with all edges that have it as an endpoint. For each integer \(k\) from \(1\) to \(n\), find the number, modulo \(998\,244\,353\), of different sequences \(a_1, a_2, \ldots, a_n\) that satisfy the following conditions: it is possible to obtain \(a\) by performing the aforementioned operations exactly \(n\) times in some order. \(\operatorname{gcd}(a_1, a_2, \ldots, a_n) = k\). Here, \(\operatorname{gcd}\) means the greatest common divisor of the elements in \(a\).
The first line contains a single integer \(t\) (\(1 \le t \le 10\,000\)) — the number of test cases.The first line of each test case contains a single integer \(n\) (\(2 \le n \le 10^5\)).Each of the next \(n - 1\) lines contains two integers \(u\) and \(v\) (\(1 \le u, v \le n\)) indicating there is an edge between vertices \(u\) and \(v\). It is guaranteed that the given edges form a tree.It is guaranteed that the sum of \(n\) over all test cases doesn't exceed \(3 \cdot 10^5\).
For each test case, print \(n\) integers in a single line, where for each \(k\) from \(1\) to \(n\), the \(k\)-th integer denotes the answer when \(\operatorname{gcd}\) equals to \(k\).
In the first test case, If we delete the nodes in order \(1 \rightarrow 2 \rightarrow 3\) or \(1 \rightarrow 3 \rightarrow 2\), then the obtained sequence will be \(a = [2, 0, 0]\) which has \(\operatorname{gcd}\) equals to \(2\). If we delete the nodes in order \(2 \rightarrow 1 \rightarrow 3\), then the obtained sequence will be \(a = [1, 1, 0]\) which has \(\operatorname{gcd}\) equals to \(1\). If we delete the nodes in order \(3 \rightarrow 1 \rightarrow 2\), then the obtained sequence will be \(a = [1, 0, 1]\) which has \(\operatorname{gcd}\) equals to \(1\). If we delete the nodes in order \(2 \rightarrow 3 \rightarrow 1\) or \(3 \rightarrow 2 \rightarrow 1\), then the obtained sequence will be \(a = [0, 1, 1]\) which has \(\operatorname{gcd}\) equals to \(1\). Note that here we are counting the number of different sequences, not the number of different orders of deleting nodes.