#include <bits/stdc++.h>
using namespace std;
vector<long long> a;
bool valido;
vector<vector<long long>> adj;
long long d;
void dfs(long long u, long long p = 0) {
  if (!valido) return;
  for (long long v : adj[u])
    if (v != p) dfs(v, u);
  if (a[u] % d == 0) {
    a[p]++;
  } else {
    a[u]++;
    if (p == 0 || a[u] % d != 0) valido = 0;
  }
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    a.resize(n + 1);
    adj = vector<vector<long long>>(n + 1);
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    vector<long long> freq(n + 1);
    freq[1] = 2;
    for (long long i = 1; i < n - 1; i++) freq[1] = 2 * freq[1] % 998244353;
    for (d = 2; d <= n - 1; d++) {
      if ((n - 1) % d == 0) {
        valido = 1;
        fill(a.begin(), a.end(), 0);
        dfs(1);
        freq[d] = valido;
      }
    }
    vector<long long> res(n + 1);
    for (long long d = n; d >= 1; d--) {
      res[d] = freq[d];
      for (long long mult = 2 * d; mult <= n; mult += d) {
        res[d] = res[d] - res[mult] + 998244353;
        res[d] %= 998244353;
      }
    }
    for (long long d = 1; d <= n; d++) {
      cout << res[d] << ' ';
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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, …, 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, …, 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, …, a_n) = k. Here, \operatorname{gcd} means the greatest common divisor of the elements in a. 

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).

Each of the next n - 1 lines contains two integers u and v (1 ≤ u, v ≤ 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 ⋅ 10^5.

Output

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.

Example

Input


2
3
2 1
1 3
2
1 2


Output


3 1 0
2 0

Note

In the first test case, 

<image>

  * If we delete the nodes in order 1 → 2 → 3 or 1 → 3 → 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 → 1 → 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 → 1 → 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 → 3 → 1 or 3 → 2 → 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<long long> a;
bool valido;
vector<vector<long long>> adj;
long long d;
void dfs(long long u, long long p = 0) {
  if (!valido) return;
  for (long long v : adj[u])
    if (v != p) dfs(v, u);
  if (a[u] % d == 0) {
    a[p]++;
  } else {
    a[u]++;
    if (p == 0 || a[u] % d != 0) valido = 0;
  }
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    a.resize(n + 1);
    adj = vector<vector<long long>>(n + 1);
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    vector<long long> freq(n + 1);
    freq[1] = 2;
    for (long long i = 1; i < n - 1; i++) freq[1] = 2 * freq[1] % 998244353;
    for (d = 2; d <= n - 1; d++) {
      if ((n - 1) % d == 0) {
        valido = 1;
        fill(a.begin(), a.end(), 0);
        dfs(1);
        freq[d] = valido;
      }
    }
    vector<long long> res(n + 1);
    for (long long d = n; d >= 1; d--) {
      res[d] = freq[d];
      for (long long mult = 2 * d; mult <= n; mult += d) {
        res[d] = res[d] - res[mult] + 998244353;
        res[d] %= 998244353;
      }
    }
    for (long long d = 1; d <= n; d++) {
      cout << res[d] << ' ';
    }
    cout << endl;
  }
  return 0;
}

```