#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long MAX = 1e5 + 5;
long long ans[MAX];
vector<long long> P[MAX];
long long pom[MAX];
bool czy = true;
long long akt;
void dfs(long long u, long long fa) {
  for (auto it : P[u]) {
    if (it == fa) continue;
    dfs(it, u);
  }
  if (pom[u] % akt == 0) {
    if (fa != -1) pom[fa]++;
  } else {
    pom[u]++;
    if (pom[u] % akt != 0) czy = false;
  }
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) P[i].clear();
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    P[a].push_back(b);
    P[b].push_back(a);
  }
  ans[1] = 1;
  for (long long i = 1; i <= n - 1; i++) ans[1] = (ans[1] * 2) % mod;
  for (long long i = 2; i <= n; i++) {
    if ((n - 1) % i == 0) {
      czy = true;
      akt = i;
      for (long long j = 1; j <= n; j++) pom[j] = 0;
      dfs(1, -1);
      ans[i] = czy;
    } else
      ans[i] = 0;
  }
  for (long long i = n; i >= 1; i--) {
    for (long long j = 2 * i; j <= n; j += i)
      ans[i] = (ans[i] - ans[j] + mod) % mod;
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long z = 0; z < t; z++) solve();
  return 0;
}

### Prompt

Please create a solution in cpp 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 long long mod = 998244353;
const long long MAX = 1e5 + 5;
long long ans[MAX];
vector<long long> P[MAX];
long long pom[MAX];
bool czy = true;
long long akt;
void dfs(long long u, long long fa) {
  for (auto it : P[u]) {
    if (it == fa) continue;
    dfs(it, u);
  }
  if (pom[u] % akt == 0) {
    if (fa != -1) pom[fa]++;
  } else {
    pom[u]++;
    if (pom[u] % akt != 0) czy = false;
  }
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) P[i].clear();
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    P[a].push_back(b);
    P[b].push_back(a);
  }
  ans[1] = 1;
  for (long long i = 1; i <= n - 1; i++) ans[1] = (ans[1] * 2) % mod;
  for (long long i = 2; i <= n; i++) {
    if ((n - 1) % i == 0) {
      czy = true;
      akt = i;
      for (long long j = 1; j <= n; j++) pom[j] = 0;
      dfs(1, -1);
      ans[i] = czy;
    } else
      ans[i] = 0;
  }
  for (long long i = n; i >= 1; i--) {
    for (long long j = 2 * i; j <= n; j += i)
      ans[i] = (ans[i] - ans[j] + mod) % mod;
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long z = 0; z < t; z++) solve();
  return 0;
}

```