#include <bits/stdc++.h>
using namespace std;
const int maxn = 300000 + 50;
const long long mm = 998244353;
int n, m, ans, T;
long long f[maxn], w[maxn];
vector<int> tr[maxn];
long long dfs1(int x, int fa) {
  long long tot = 1;
  w[x] = 1;
  for (int i = 0; i < tr[x].size(); ++i)
    if (tr[x][i] != fa)
      tot = tot * dfs1(tr[x][i], x) * 2 % mm, w[x] = (w[x] + w[tr[x][i]]);
  return tot;
}
int dfs(int x, int fa, int z, int y) {
  int tot = 0;
  for (int i = 0; i < tr[x].size(); ++i)
    if (tr[x][i] != fa)
      if (w[tr[x][i]] % z == 0) {
        if (!dfs(tr[x][i], x, z, 1)) return 0;
      } else if ((w[tr[x][i]] - 1) % z == 0) {
        ++tot;
        if (!dfs(tr[x][i], x, z, 0)) return 0;
      } else
        return 0;
  if ((tot + y) % z == 0)
    ;
  else
    return 0;
  return 1;
}
int main() {
  scanf("%d", &T);
  for (int oo = 1; oo <= T; ++oo) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
      while (!tr[i].empty()) tr[i].pop_back();
    for (int i = 1; i < n; ++i) {
      int q1, q2;
      scanf("%d%d", &q1, &q2);
      tr[q1].push_back(q2);
      tr[q2].push_back(q1);
    }
    for (int i = 1; i <= n; ++i) f[i] = 0;
    f[1] = dfs1(1, 1);
    for (int i = n - 1; i > 1; --i)
      if ((n - 1) % i == 0 && !f[i]) {
        f[i] += dfs(1, 1, i, 0);
        if (f[i])
          for (int j = 1; j * j <= i; ++j)
            if (i % j == 0) {
              f[j] -= f[i];
              if (j * j != i && j != 1) f[i / j] -= f[i];
            }
      }
    for (int i = 1; i <= n; ++i)
      if (f[i] >= 0)
        printf("%lld ", (f[i] + mm) % mm);
      else
        printf("0 ");
    printf("\n");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 maxn = 300000 + 50;
const long long mm = 998244353;
int n, m, ans, T;
long long f[maxn], w[maxn];
vector<int> tr[maxn];
long long dfs1(int x, int fa) {
  long long tot = 1;
  w[x] = 1;
  for (int i = 0; i < tr[x].size(); ++i)
    if (tr[x][i] != fa)
      tot = tot * dfs1(tr[x][i], x) * 2 % mm, w[x] = (w[x] + w[tr[x][i]]);
  return tot;
}
int dfs(int x, int fa, int z, int y) {
  int tot = 0;
  for (int i = 0; i < tr[x].size(); ++i)
    if (tr[x][i] != fa)
      if (w[tr[x][i]] % z == 0) {
        if (!dfs(tr[x][i], x, z, 1)) return 0;
      } else if ((w[tr[x][i]] - 1) % z == 0) {
        ++tot;
        if (!dfs(tr[x][i], x, z, 0)) return 0;
      } else
        return 0;
  if ((tot + y) % z == 0)
    ;
  else
    return 0;
  return 1;
}
int main() {
  scanf("%d", &T);
  for (int oo = 1; oo <= T; ++oo) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
      while (!tr[i].empty()) tr[i].pop_back();
    for (int i = 1; i < n; ++i) {
      int q1, q2;
      scanf("%d%d", &q1, &q2);
      tr[q1].push_back(q2);
      tr[q2].push_back(q1);
    }
    for (int i = 1; i <= n; ++i) f[i] = 0;
    f[1] = dfs1(1, 1);
    for (int i = n - 1; i > 1; --i)
      if ((n - 1) % i == 0 && !f[i]) {
        f[i] += dfs(1, 1, i, 0);
        if (f[i])
          for (int j = 1; j * j <= i; ++j)
            if (i % j == 0) {
              f[j] -= f[i];
              if (j * j != i && j != 1) f[i / j] -= f[i];
            }
      }
    for (int i = 1; i <= n; ++i)
      if (f[i] >= 0)
        printf("%lld ", (f[i] + mm) % mm);
      else
        printf("0 ");
    printf("\n");
  }
  return 0;
}

```