#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
const int MAXN = 3e5 + 11;
vector<int> vec[MAXN];
int cas, N, Ans[MAXN], nw[MAXN];
bool ff;
void dfs(int u, int fath, int Lim) {
  nw[u] = 0;
  for (auto v : vec[u])
    if (v != fath) dfs(v, u, Lim);
  if (fath) {
    if (nw[u])
      nw[u] = (nw[u] + 1) % Lim;
    else
      nw[fath] = (nw[fath] + 1) % Lim;
  }
  if (nw[u]) ff = 0;
  return;
}
signed main() {
  cas = read();
  while (cas--) {
    N = read();
    for (int i = 1; i <= N; i++) vec[i].clear(), Ans[i] = 0;
    int res = 1;
    for (int i = 1; i < N; i++) {
      res = res * 2 % 998244353;
      int u = read(), v = read();
      vec[u].push_back(v), vec[v].push_back(u);
    }
    int cnt = 0;
    for (int i = 2; i <= N; i++)
      if (!((N - 1) % i)) {
        ff = 1;
        dfs(1, 0, i);
        Ans[i] = ff;
      }
    for (int i = N - 1; i >= 2; i--)
      for (int j = i + i; j <= N - 1; j += i) Ans[i] -= Ans[j];
    for (int i = 2; i <= N - 1; i++) cnt += Ans[i];
    Ans[1] = (res - cnt + 998244353) % 998244353;
    for (int i = 1; i <= N; i++) printf("%d ", Ans[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
inline int read() {
  int f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
const int MAXN = 3e5 + 11;
vector<int> vec[MAXN];
int cas, N, Ans[MAXN], nw[MAXN];
bool ff;
void dfs(int u, int fath, int Lim) {
  nw[u] = 0;
  for (auto v : vec[u])
    if (v != fath) dfs(v, u, Lim);
  if (fath) {
    if (nw[u])
      nw[u] = (nw[u] + 1) % Lim;
    else
      nw[fath] = (nw[fath] + 1) % Lim;
  }
  if (nw[u]) ff = 0;
  return;
}
signed main() {
  cas = read();
  while (cas--) {
    N = read();
    for (int i = 1; i <= N; i++) vec[i].clear(), Ans[i] = 0;
    int res = 1;
    for (int i = 1; i < N; i++) {
      res = res * 2 % 998244353;
      int u = read(), v = read();
      vec[u].push_back(v), vec[v].push_back(u);
    }
    int cnt = 0;
    for (int i = 2; i <= N; i++)
      if (!((N - 1) % i)) {
        ff = 1;
        dfs(1, 0, i);
        Ans[i] = ff;
      }
    for (int i = N - 1; i >= 2; i--)
      for (int j = i + i; j <= N - 1; j += i) Ans[i] -= Ans[j];
    for (int i = 2; i <= N - 1; i++) cnt += Ans[i];
    Ans[1] = (res - cnt + 998244353) % 998244353;
    for (int i = 1; i <= N; i++) printf("%d ", Ans[i]);
    printf("\n");
  }
  return 0;
}

```