#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> adjList[300005];
vector<int> visited(300005);
vector<int> a(300005);
vector<long long> f(300005);
vector<long long> twoPowers(300005);
long long mod = 998244353;
vector<long long> ans(300005);
int dfs(int node, int goal) {
  visited[node] = 1;
  int accrued = 0;
  for (int i = 0; i < adjList[node].size(); i++) {
    int child = adjList[node][i];
    if (!visited[child]) {
      int au = dfs(child, goal);
      if (au == -1) return -1;
      accrued += au;
    }
  }
  int toReturn = 1;
  if (accrued % goal == 0)
    toReturn = 1;
  else if ((accrued + 1) % goal == 0)
    toReturn = 0;
  else
    toReturn = -1;
  return toReturn;
}
int main() {
  iostream::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  twoPowers[0] = 1;
  for (int i = 1; i < 300005; i++) {
    twoPowers[i] = (twoPowers[i - 1] * 2) % mod;
  }
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n + 1; i++) {
      vector<int> v;
      adjList[i] = v;
      f[i] = 0;
      ans[i] = 0;
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      adjList[u].push_back(v);
      adjList[v].push_back(u);
    }
    f[1] = twoPowers[n - 1];
    for (int i = 2; i <= n - 1; i++) {
      if ((n - 1) % i == 0) {
        for (int j = 0; j <= n; j++) {
          visited[j] = 0;
          a[j] = 0;
        }
        int res = dfs(0, i);
        if (res == 1) f[i] = 1;
      }
    }
    for (int i = n - 1; i >= 1; i--) {
      ans[i] = f[i];
      for (int j = 2; j * i <= n - 1; j++) ans[i] = ans[i] - ans[j * i];
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
    cout << endl;
  }
}

### Prompt

Your task is to create 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;
int n;
vector<int> adjList[300005];
vector<int> visited(300005);
vector<int> a(300005);
vector<long long> f(300005);
vector<long long> twoPowers(300005);
long long mod = 998244353;
vector<long long> ans(300005);
int dfs(int node, int goal) {
  visited[node] = 1;
  int accrued = 0;
  for (int i = 0; i < adjList[node].size(); i++) {
    int child = adjList[node][i];
    if (!visited[child]) {
      int au = dfs(child, goal);
      if (au == -1) return -1;
      accrued += au;
    }
  }
  int toReturn = 1;
  if (accrued % goal == 0)
    toReturn = 1;
  else if ((accrued + 1) % goal == 0)
    toReturn = 0;
  else
    toReturn = -1;
  return toReturn;
}
int main() {
  iostream::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  twoPowers[0] = 1;
  for (int i = 1; i < 300005; i++) {
    twoPowers[i] = (twoPowers[i - 1] * 2) % mod;
  }
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n + 1; i++) {
      vector<int> v;
      adjList[i] = v;
      f[i] = 0;
      ans[i] = 0;
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      adjList[u].push_back(v);
      adjList[v].push_back(u);
    }
    f[1] = twoPowers[n - 1];
    for (int i = 2; i <= n - 1; i++) {
      if ((n - 1) % i == 0) {
        for (int j = 0; j <= n; j++) {
          visited[j] = 0;
          a[j] = 0;
        }
        int res = dfs(0, i);
        if (res == 1) f[i] = 1;
      }
    }
    for (int i = n - 1; i >= 1; i--) {
      ans[i] = f[i];
      for (int j = 2; j * i <= n - 1; j++) ans[i] = ans[i] - ans[j * i];
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
    cout << endl;
  }
}

```