#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 7;
const int MOD = 998244353;
int t;
int n;
vector<int> G[MAX];
int f[MAX], k;
bool dfs(int u, int fa) {
  f[u] = 0;
  if (u != 1 && G[u].size() == 1) return true;
  for (auto v : G[u]) {
    if (v == fa) continue;
    if (dfs(v, u)) f[u]++;
  }
  if (fa && f[u] % k != 0) {
    f[u]++;
    return false;
  }
  return true;
}
int qw[MAX];
int ans[MAX];
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  qw[0] = 1;
  for (int i = 1; i <= 100000; i++) qw[i] = qw[i - 1] * 2 % MOD;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) G[i].clear(), ans[i] = 0;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      G[u].push_back(v), G[v].push_back(u);
    }
    ans[1] = qw[n - 1];
    for (int i = 2; i < n; i++) {
      if ((n - 1) % i != 0) continue;
      k = i;
      dfs(1, 0);
      int fg = 0;
      for (int j = 1; j <= n; j++)
        if (f[j] % i) {
          fg = 1;
          break;
        }
      if (!fg) ans[i]++;
    }
    for (int i = n; i >= 1; i--) {
      for (int j = 2 * i; j <= n; j += i) ans[i] -= ans[j];
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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;
const int MAX = 1e5 + 7;
const int MOD = 998244353;
int t;
int n;
vector<int> G[MAX];
int f[MAX], k;
bool dfs(int u, int fa) {
  f[u] = 0;
  if (u != 1 && G[u].size() == 1) return true;
  for (auto v : G[u]) {
    if (v == fa) continue;
    if (dfs(v, u)) f[u]++;
  }
  if (fa && f[u] % k != 0) {
    f[u]++;
    return false;
  }
  return true;
}
int qw[MAX];
int ans[MAX];
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  qw[0] = 1;
  for (int i = 1; i <= 100000; i++) qw[i] = qw[i - 1] * 2 % MOD;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) G[i].clear(), ans[i] = 0;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      G[u].push_back(v), G[v].push_back(u);
    }
    ans[1] = qw[n - 1];
    for (int i = 2; i < n; i++) {
      if ((n - 1) % i != 0) continue;
      k = i;
      dfs(1, 0);
      int fg = 0;
      for (int j = 1; j <= n; j++)
        if (f[j] % i) {
          fg = 1;
          break;
        }
      if (!fg) ans[i]++;
    }
    for (int i = n; i >= 1; i--) {
      for (int j = 2 * i; j <= n; j += i) ans[i] -= ans[j];
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << '\n';
  }
  return 0;
}

```