#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int P = 998244353;
int T, n;
int ans[N];
vector<int> e[N];
int f[N];
bool dfs(int x, int fa, int y) {
  for (auto v : e[x])
    if (v != fa && !dfs(v, x, y)) return 0;
  if (f[x] % y == 0) {
    f[fa]--;
    return 1;
  }
  f[x]--;
  if (f[x] % y == 0) return 1;
  return 0;
}
vector<int> pri;
bool is[N];
int mu[N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cin >> T;
  for (int i = 2; i < N; i++) {
    if (!is[i]) pri.push_back(i), mu[i] = -1;
    for (int j = 0; j < ((int)pri.size()) && i * pri[j] < N; j++) {
      is[i * pri[j]] = 1;
      mu[i * pri[j]] = i % pri[j] ? -mu[i] : 0;
      if (i % pri[j] == 0) break;
    }
  }
  while (T--) {
    cin >> n;
    for (int i = 1; i <= n; i++) e[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    if (n == 2) {
      cout << 2 << " " << 0 << "\n";
    } else {
      int rt = 0;
      for (int i = 1; i <= n; i++)
        if (((int)e[i].size()) > 1) rt = i;
      for (int i = 1; i <= n; i++) ans[i] = 0;
      for (int i = 1; i * i <= (n - 1); i++) {
        if ((n - 1) % i) continue;
        for (int j = 1; j <= n; j++) f[j] = ((int)e[j].size());
        if (dfs(rt, 0, i)) ans[i] = 1;
        for (int j = 1; j <= n; j++) f[j] = ((int)e[j].size());
        if (dfs(rt, 0, (n - 1) / i)) ans[(n - 1) / i] = 1;
      }
      for (int i = 2; i <= n; i++) {
        for (int j = i + i; j <= n; j += i)
          ans[i] = (ans[i] + mu[j / i] * ans[j]);
        ans[i] = (ans[i] + P) % P;
      }
      for (int i = 1; i < n; i++) ans[1] = 2 * ans[1] % P;
      for (int i = 2; i <= n; i++) ans[1] = (ans[1] - ans[i] + P) % P;
      for (int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
    }
  }
  return 0;
}

### Prompt

Please formulate 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 + 10;
const int P = 998244353;
int T, n;
int ans[N];
vector<int> e[N];
int f[N];
bool dfs(int x, int fa, int y) {
  for (auto v : e[x])
    if (v != fa && !dfs(v, x, y)) return 0;
  if (f[x] % y == 0) {
    f[fa]--;
    return 1;
  }
  f[x]--;
  if (f[x] % y == 0) return 1;
  return 0;
}
vector<int> pri;
bool is[N];
int mu[N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cin >> T;
  for (int i = 2; i < N; i++) {
    if (!is[i]) pri.push_back(i), mu[i] = -1;
    for (int j = 0; j < ((int)pri.size()) && i * pri[j] < N; j++) {
      is[i * pri[j]] = 1;
      mu[i * pri[j]] = i % pri[j] ? -mu[i] : 0;
      if (i % pri[j] == 0) break;
    }
  }
  while (T--) {
    cin >> n;
    for (int i = 1; i <= n; i++) e[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    if (n == 2) {
      cout << 2 << " " << 0 << "\n";
    } else {
      int rt = 0;
      for (int i = 1; i <= n; i++)
        if (((int)e[i].size()) > 1) rt = i;
      for (int i = 1; i <= n; i++) ans[i] = 0;
      for (int i = 1; i * i <= (n - 1); i++) {
        if ((n - 1) % i) continue;
        for (int j = 1; j <= n; j++) f[j] = ((int)e[j].size());
        if (dfs(rt, 0, i)) ans[i] = 1;
        for (int j = 1; j <= n; j++) f[j] = ((int)e[j].size());
        if (dfs(rt, 0, (n - 1) / i)) ans[(n - 1) / i] = 1;
      }
      for (int i = 2; i <= n; i++) {
        for (int j = i + i; j <= n; j += i)
          ans[i] = (ans[i] + mu[j / i] * ans[j]);
        ans[i] = (ans[i] + P) % P;
      }
      for (int i = 1; i < n; i++) ans[1] = 2 * ans[1] % P;
      for (int i = 2; i <= n; i++) ans[1] = (ans[1] - ans[i] + P) % P;
      for (int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
    }
  }
  return 0;
}

```