#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
const int base = 998244353;
int t, n, f[N], d[N], par[N];
vector<int> a[N];
void dfs(int u, int p, int c) {
  if (f[c] == 0) return;
  d[u] = 0;
  for (auto v : a[u])
    if (v != p) dfs(v, u, c);
  if (d[u] % c == 0)
    d[p]++;
  else if ((d[u] + 1) % c != 0)
    f[c] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  for (int iTest = 1; iTest <= t; iTest++) {
    cin >> n;
    for (int i = 1; i <= n; i++) a[i].clear(), f[i] = 0;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      a[u].push_back(v);
      a[v].push_back(u);
    }
    f[n] = 0;
    f[1] = 1;
    for (int i = 1; i < n; i++) f[1] = (f[1] * 2) % base;
    for (int i = 2; i < n; i++)
      if ((n - 1) % i == 0) {
        f[i] = 1;
        dfs(1, 0, i);
      }
    for (int i = n - 1; i > 1; i--)
      for (int j = i * 2; j < n; j += i)
        if (f[j] > 0) {
          f[i] = 0;
          break;
        }
    for (int i = 2; i < n; i++) f[1] = (f[1] - f[i] + base) % base;
    for (int i = 1; i <= n; i++) cout << f[i] << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Generate 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 N = 300005;
const int base = 998244353;
int t, n, f[N], d[N], par[N];
vector<int> a[N];
void dfs(int u, int p, int c) {
  if (f[c] == 0) return;
  d[u] = 0;
  for (auto v : a[u])
    if (v != p) dfs(v, u, c);
  if (d[u] % c == 0)
    d[p]++;
  else if ((d[u] + 1) % c != 0)
    f[c] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  for (int iTest = 1; iTest <= t; iTest++) {
    cin >> n;
    for (int i = 1; i <= n; i++) a[i].clear(), f[i] = 0;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      a[u].push_back(v);
      a[v].push_back(u);
    }
    f[n] = 0;
    f[1] = 1;
    for (int i = 1; i < n; i++) f[1] = (f[1] * 2) % base;
    for (int i = 2; i < n; i++)
      if ((n - 1) % i == 0) {
        f[i] = 1;
        dfs(1, 0, i);
      }
    for (int i = n - 1; i > 1; i--)
      for (int j = i * 2; j < n; j += i)
        if (f[j] > 0) {
          f[i] = 0;
          break;
        }
    for (int i = 2; i < n; i++) f[1] = (f[1] - f[i] + base) % base;
    for (int i = 1; i <= n; i++) cout << f[i] << " ";
    cout << "\n";
  }
  return 0;
}

```