#include <bits/stdc++.h>
using namespace std;
long long M = 998244353;
long long fpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long p = fpow(x, y / 2);
  p *= p;
  p %= M;
  if (y & 1) p *= x;
  return p % M;
}
vector<int> v[100005];
long long ans[100005], po[100005];
bool ok;
int dfs(int ver, int k, int par) {
  int sum = 0;
  for (int i = 0; i < v[ver].size(); i++) {
    if (v[ver][i] == par) continue;
    sum += dfs(v[ver][i], k, ver);
  }
  if (sum % k == 0) {
    return 1;
  }
  if ((sum + 1) % k != 0) ok = 0;
  return 0;
}
int main() {
  po[0] = 1;
  for (int i = 1; i < 100005; i++) {
    po[i] = 2 * po[i - 1];
    po[i] %= M;
  }
  int t, x, y, n;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    ans[1] = po[n - 1];
    for (int k = 2; k <= n; k++) {
      ok = 1;
      ans[k] = 0;
      if ((n - 1) % k != 0) continue;
      dfs(1, k, -1);
      ans[k] = ok;
    }
    for (int k = n; k >= 1; k--) {
      for (int i = k * 2; i <= n; i += k) {
        ans[k] = (ans[k] - ans[i] + M) % M;
      }
    }
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << ' ';
      v[i].clear();
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long M = 998244353;
long long fpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long p = fpow(x, y / 2);
  p *= p;
  p %= M;
  if (y & 1) p *= x;
  return p % M;
}
vector<int> v[100005];
long long ans[100005], po[100005];
bool ok;
int dfs(int ver, int k, int par) {
  int sum = 0;
  for (int i = 0; i < v[ver].size(); i++) {
    if (v[ver][i] == par) continue;
    sum += dfs(v[ver][i], k, ver);
  }
  if (sum % k == 0) {
    return 1;
  }
  if ((sum + 1) % k != 0) ok = 0;
  return 0;
}
int main() {
  po[0] = 1;
  for (int i = 1; i < 100005; i++) {
    po[i] = 2 * po[i - 1];
    po[i] %= M;
  }
  int t, x, y, n;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    ans[1] = po[n - 1];
    for (int k = 2; k <= n; k++) {
      ok = 1;
      ans[k] = 0;
      if ((n - 1) % k != 0) continue;
      dfs(1, k, -1);
      ans[k] = ok;
    }
    for (int k = n; k >= 1; k--) {
      for (int i = k * 2; i <= n; i += k) {
        ans[k] = (ans[k] - ans[i] + M) % M;
      }
    }
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << ' ';
      v[i].clear();
    }
    cout << endl;
  }
  return 0;
}

```