#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const int MONKE = 0;
const ll mod = 998244353;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> edges(n - 1);
  vector<vector<int>> adj(n);
  for (auto &[l, r] : edges) {
    cin >> l >> r;
    adj[l - 1].push_back(r - 1);
    adj[r - 1].push_back(l - 1);
  }
  vector<ll> f(n + 1);
  f[1] = 1;
  for (int i = 1; i < n; i++) f[1] = (f[1] + f[1]) % mod;
  function<int(int, int, int)> dfs = [&](int node, int par, int val) {
    int cnt = 0;
    for (auto e : adj[node]) {
      if (e == par) continue;
      int res = dfs(e, node, val);
      if (res == -1) return -1;
      cnt += res;
    }
    if (cnt % val == 0) return 1;
    if (cnt % val == val - 1) return 0;
    return -1;
  };
  for (int i = 2; i < n; i++) {
    if ((n - 1) % i == 0) {
      f[i] = (dfs(0, -1, i) > 0);
    }
  }
  for (int i = n; i > 0; i--) {
    for (int j = i + i; j < n; j += i) f[i] -= f[j];
    if (i == 1) f[i] = (f[i] + mod) % mod;
  }
  for (int i = 1; i <= n; i++) cout << f[i] << " \n"[i == n];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (int tc = 1; tc <= t; tc++) {
    solve();
  }
  return MONKE;
}

### 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>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const int MONKE = 0;
const ll mod = 998244353;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> edges(n - 1);
  vector<vector<int>> adj(n);
  for (auto &[l, r] : edges) {
    cin >> l >> r;
    adj[l - 1].push_back(r - 1);
    adj[r - 1].push_back(l - 1);
  }
  vector<ll> f(n + 1);
  f[1] = 1;
  for (int i = 1; i < n; i++) f[1] = (f[1] + f[1]) % mod;
  function<int(int, int, int)> dfs = [&](int node, int par, int val) {
    int cnt = 0;
    for (auto e : adj[node]) {
      if (e == par) continue;
      int res = dfs(e, node, val);
      if (res == -1) return -1;
      cnt += res;
    }
    if (cnt % val == 0) return 1;
    if (cnt % val == val - 1) return 0;
    return -1;
  };
  for (int i = 2; i < n; i++) {
    if ((n - 1) % i == 0) {
      f[i] = (dfs(0, -1, i) > 0);
    }
  }
  for (int i = n; i > 0; i--) {
    for (int j = i + i; j < n; j += i) f[i] -= f[j];
    if (i == 1) f[i] = (f[i] + mod) % mod;
  }
  for (int i = 1; i <= n; i++) cout << f[i] << " \n"[i == n];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (int tc = 1; tc <= t; tc++) {
    solve();
  }
  return MONKE;
}

```