#include <bits/stdc++.h>
using namespace std;
long long t;
vector<long long> adj[100001];
long long ss[100001];
void dfs(long long no, long long par = -1) {
  ss[no] = 1;
  for (auto j : adj[no]) {
    if (j != par) {
      dfs(j, no);
      ss[no] += ss[j];
    }
  }
}
const long long mod = 998244353;
long long ans[100001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < n; i++) {
      adj[i].clear();
    }
    for (long long i = 0; i < n - 1; i++) {
      long long aa, bb;
      cin >> aa >> bb;
      aa--;
      bb--;
      adj[aa].push_back(bb);
      adj[bb].push_back(aa);
    }
    dfs(0);
    vector<pair<long long, long long>> ss2;
    vector<long long> tt;
    for (long long i = 1; i <= n; i++) {
      ans[i] = 0;
    }
    for (long long i = 1; i < n; i++) {
      ss2.push_back({ss[i] - 1, ss[i]});
    }
    for (long long i = 1; i <= n - 1; i++) {
      if ((n - 1) % i == 0) {
        tt.push_back(i);
      }
    }
    reverse(tt.begin(), tt.end());
    for (auto j : tt) {
      ans[j] = 1;
      for (auto k : ss2) {
        long long co = 0;
        if ((k.first) % j == 0) {
          co++;
        }
        if ((k.second) % j == 0) {
          co++;
        }
        ans[j] = (ans[j] * co) % mod;
      }
      for (auto i : tt) {
        if (i > j) {
          if ((i % j) == 0) {
            ans[j] = (ans[j] - ans[i] + mod) % mod;
          }
        }
      }
    }
    for (long long 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;
long long t;
vector<long long> adj[100001];
long long ss[100001];
void dfs(long long no, long long par = -1) {
  ss[no] = 1;
  for (auto j : adj[no]) {
    if (j != par) {
      dfs(j, no);
      ss[no] += ss[j];
    }
  }
}
const long long mod = 998244353;
long long ans[100001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < n; i++) {
      adj[i].clear();
    }
    for (long long i = 0; i < n - 1; i++) {
      long long aa, bb;
      cin >> aa >> bb;
      aa--;
      bb--;
      adj[aa].push_back(bb);
      adj[bb].push_back(aa);
    }
    dfs(0);
    vector<pair<long long, long long>> ss2;
    vector<long long> tt;
    for (long long i = 1; i <= n; i++) {
      ans[i] = 0;
    }
    for (long long i = 1; i < n; i++) {
      ss2.push_back({ss[i] - 1, ss[i]});
    }
    for (long long i = 1; i <= n - 1; i++) {
      if ((n - 1) % i == 0) {
        tt.push_back(i);
      }
    }
    reverse(tt.begin(), tt.end());
    for (auto j : tt) {
      ans[j] = 1;
      for (auto k : ss2) {
        long long co = 0;
        if ((k.first) % j == 0) {
          co++;
        }
        if ((k.second) % j == 0) {
          co++;
        }
        ans[j] = (ans[j] * co) % mod;
      }
      for (auto i : tt) {
        if (i > j) {
          if ((i % j) == 0) {
            ans[j] = (ans[j] - ans[i] + mod) % mod;
          }
        }
      }
    }
    for (long long i = 1; i <= n; i++) {
      cout << ans[i] << " ";
    }
    cout << '\n';
  }
  return 0;
}

```