#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10, MOD = 998244353;
int n;
vector<int> adj[MAXN];
bool work[MAXN], ans[MAXN];
int dfs(int c, int p, int k) {
  int me = 0;
  for (auto nxt : adj[c])
    if (nxt != p) {
      int they = dfs(nxt, c, k);
      if (they == -1) return -1;
      assert(they == 0 || they == 1);
      me += they;
    }
  if ((me + 1) % k == 0) return 0;
  if (me % k == 0) return 1;
  return -1;
}
void solve() {
  cin >> n;
  for (int i = 0; i <= n; i++) adj[i].clear(), ans[i] = work[i] = 0;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b, --a, --b;
    adj[a].push_back(b), adj[b].push_back(a);
  }
  for (int i = 2; i <= n; i++)
    if ((n - 1) % i == 0) {
      work[i] = dfs(0, -1, i) == 1;
    }
  int cur = 1;
  for (int i = 0; i < n - 1; i++) cur = cur * 2 % MOD;
  for (int i = 2; i <= n; i++) {
    ans[i] = work[i];
    for (int j = 2 * i; j <= n; j += i) ans[i] &= !work[j];
    cur -= ans[i];
  }
  for (int i = 1; i <= n; i++) {
    if (i == 1)
      cout << cur;
    else
      cout << ' ' << ans[i];
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = 1;
  cin >> T;
  while (T--) solve();
}

### Prompt

Create a solution in Cpp for 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 = 1e5 + 10, MOD = 998244353;
int n;
vector<int> adj[MAXN];
bool work[MAXN], ans[MAXN];
int dfs(int c, int p, int k) {
  int me = 0;
  for (auto nxt : adj[c])
    if (nxt != p) {
      int they = dfs(nxt, c, k);
      if (they == -1) return -1;
      assert(they == 0 || they == 1);
      me += they;
    }
  if ((me + 1) % k == 0) return 0;
  if (me % k == 0) return 1;
  return -1;
}
void solve() {
  cin >> n;
  for (int i = 0; i <= n; i++) adj[i].clear(), ans[i] = work[i] = 0;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b, --a, --b;
    adj[a].push_back(b), adj[b].push_back(a);
  }
  for (int i = 2; i <= n; i++)
    if ((n - 1) % i == 0) {
      work[i] = dfs(0, -1, i) == 1;
    }
  int cur = 1;
  for (int i = 0; i < n - 1; i++) cur = cur * 2 % MOD;
  for (int i = 2; i <= n; i++) {
    ans[i] = work[i];
    for (int j = 2 * i; j <= n; j += i) ans[i] &= !work[j];
    cur -= ans[i];
  }
  for (int i = 1; i <= n; i++) {
    if (i == 1)
      cout << cur;
    else
      cout << ' ' << ans[i];
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = 1;
  cin >> T;
  while (T--) solve();
}

```