#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long int gcd(long long int A, long long int B) {
  if (!B) return A;
  return gcd(B, A % B);
}
vector<vector<int>> adj(100005);
vector<int> dp(100005, 0);
bool dfs(int v, int u, int k) {
  int cur = 0;
  for (int x : adj[v]) {
    if (x == u) continue;
    cur += dfs(x, v, k);
  }
  bool res = 0;
  if (cur % k && u != -1) {
    cur++;
    res = 1;
  }
  if (cur % k) dp[k] = 0;
  return res ^ 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      adj[i].clear();
      dp[i + 1] = 0;
    }
    for (int i = 0; i < n - 1; i++) {
      int v, u;
      cin >> v >> u;
      v--;
      u--;
      adj[v].push_back(u);
      adj[u].push_back(v);
    }
    n--;
    vector<int> fac = {1};
    dp[1] = 1;
    for (int i = 0; i < n; i++) {
      dp[1] *= 2;
      if (dp[1] >= (long long int)(998244353))
        dp[1] -= (long long int)(998244353);
    }
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        dp[i] = 1;
        dfs(0, -1, i);
        fac.push_back(i);
        if (i * i != n) {
          dp[n / i] = 1;
          dfs(0, -1, n / i);
          fac.push_back(n / i);
        }
      }
    }
    if (n != 1) {
      dp[n] = 1;
      fac.push_back(n);
      dfs(0, -1, n);
    }
    sort(fac.begin(), fac.end());
    for (int i = (int)(fac.size()) - 1; i >= 0; i--) {
      for (int j = 2 * fac[i]; j <= n; j += fac[i]) {
        dp[fac[i]] -= dp[j];
      }
    }
    dp[1] += (long long int)(998244353);
    if (dp[1] >= (long long int)(998244353))
      dp[1] -= (long long int)(998244353);
    for (int i = 1; i <= n + 1; i++) cout << dp[i] << " ";
    cout << '\n';
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize("Ofast")
using namespace std;
long long int gcd(long long int A, long long int B) {
  if (!B) return A;
  return gcd(B, A % B);
}
vector<vector<int>> adj(100005);
vector<int> dp(100005, 0);
bool dfs(int v, int u, int k) {
  int cur = 0;
  for (int x : adj[v]) {
    if (x == u) continue;
    cur += dfs(x, v, k);
  }
  bool res = 0;
  if (cur % k && u != -1) {
    cur++;
    res = 1;
  }
  if (cur % k) dp[k] = 0;
  return res ^ 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      adj[i].clear();
      dp[i + 1] = 0;
    }
    for (int i = 0; i < n - 1; i++) {
      int v, u;
      cin >> v >> u;
      v--;
      u--;
      adj[v].push_back(u);
      adj[u].push_back(v);
    }
    n--;
    vector<int> fac = {1};
    dp[1] = 1;
    for (int i = 0; i < n; i++) {
      dp[1] *= 2;
      if (dp[1] >= (long long int)(998244353))
        dp[1] -= (long long int)(998244353);
    }
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        dp[i] = 1;
        dfs(0, -1, i);
        fac.push_back(i);
        if (i * i != n) {
          dp[n / i] = 1;
          dfs(0, -1, n / i);
          fac.push_back(n / i);
        }
      }
    }
    if (n != 1) {
      dp[n] = 1;
      fac.push_back(n);
      dfs(0, -1, n);
    }
    sort(fac.begin(), fac.end());
    for (int i = (int)(fac.size()) - 1; i >= 0; i--) {
      for (int j = 2 * fac[i]; j <= n; j += fac[i]) {
        dp[fac[i]] -= dp[j];
      }
    }
    dp[1] += (long long int)(998244353);
    if (dp[1] >= (long long int)(998244353))
      dp[1] -= (long long int)(998244353);
    for (int i = 1; i <= n + 1; i++) cout << dp[i] << " ";
    cout << '\n';
  }
}

```