#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9, mod = 998244353;
vector<int> g[N];
int dp[N], d, ok, ans[N];
void dfs(int u, int p = 0) {
  if (!ok) return;
  for (auto v : g[u]) {
    if (v ^ p) {
      dfs(v, u);
    }
  }
  if (dp[u] % d != 0) {
    if (p) {
      dp[u]++;
    }
    if (dp[u] % d != 0) {
      ok = 0;
      return;
    }
  } else {
    dp[p]++;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    for (int i = 1; i <= n; i++) {
      ans[i] = 0;
    }
    ans[1] = 1;
    for (int k = 1; k <= n - 1; k++) {
      ans[1] = (ans[1] + ans[1]) % mod;
    }
    for (d = 2; d <= n - 1; d++) {
      if ((n - 1) % d == 0) {
        ok = 1;
        dfs(1);
        ans[d] = ok;
        for (int i = 0; i <= n; i++) {
          dp[i] = 0;
        }
      }
    }
    for (int i = n; i >= 1; i--) {
      for (int j = i + i; j <= n; j += i) {
        ans[i] = (ans[i] - ans[j] + mod) % mod;
      }
    }
    for (int i = 1; i <= n; i++) {
      if (i > 1) cout << ' ';
      cout << ans[i];
    }
    cout << '\n';
    for (int i = 1; i <= n; i++) {
      g[i].clear();
    }
  }
  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 = 1e5 + 9, mod = 998244353;
vector<int> g[N];
int dp[N], d, ok, ans[N];
void dfs(int u, int p = 0) {
  if (!ok) return;
  for (auto v : g[u]) {
    if (v ^ p) {
      dfs(v, u);
    }
  }
  if (dp[u] % d != 0) {
    if (p) {
      dp[u]++;
    }
    if (dp[u] % d != 0) {
      ok = 0;
      return;
    }
  } else {
    dp[p]++;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    for (int i = 1; i <= n; i++) {
      ans[i] = 0;
    }
    ans[1] = 1;
    for (int k = 1; k <= n - 1; k++) {
      ans[1] = (ans[1] + ans[1]) % mod;
    }
    for (d = 2; d <= n - 1; d++) {
      if ((n - 1) % d == 0) {
        ok = 1;
        dfs(1);
        ans[d] = ok;
        for (int i = 0; i <= n; i++) {
          dp[i] = 0;
        }
      }
    }
    for (int i = n; i >= 1; i--) {
      for (int j = i + i; j <= n; j += i) {
        ans[i] = (ans[i] - ans[j] + mod) % mod;
      }
    }
    for (int i = 1; i <= n; i++) {
      if (i > 1) cout << ' ';
      cout << ans[i];
    }
    cout << '\n';
    for (int i = 1; i <= n; i++) {
      g[i].clear();
    }
  }
  return 0;
}

```