#include <bits/stdc++.h>
using namespace std;
const long long MN = 100005, MOD = 998244353;
long long t, n, u, v, val[MN], ans[MN];
vector<long long> a[MN];
void dfs(long long cur, long long par, long long k) {
  for (long long nxt : a[cur]) {
    if (nxt == par) continue;
    dfs(nxt, cur, k);
  }
  if (val[cur] % k && par)
    val[cur]++;
  else
    val[par]++;
}
int32_t main() {
  cin.tie(0), cin.sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long i = 1; i < n; i++) {
      cin >> u >> v;
      a[u].push_back(v);
      a[v].push_back(u);
    }
    ans[1] = 1;
    for (long long i = 1; i < n; i++) ans[1] = ans[1] * 2 % MOD;
    for (long long i = 2; i <= n; i++) {
      if ((n - 1) % i) continue;
      memset(val, 0, sizeof(val));
      dfs(1, 0, i);
      ans[i] = 1;
      for (long long j = 1; j <= n; j++)
        if (val[j] % i) ans[i] = 0;
    }
    for (long long i = n; i > 0; i--) {
      for (long long j = i + i; j <= n; j += i) ans[i] -= ans[j];
    }
    for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
    printf("\n");
    for (long long i = 1; i <= n; i++) a[i].clear(), val[i] = ans[i] = 0;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 long long MN = 100005, MOD = 998244353;
long long t, n, u, v, val[MN], ans[MN];
vector<long long> a[MN];
void dfs(long long cur, long long par, long long k) {
  for (long long nxt : a[cur]) {
    if (nxt == par) continue;
    dfs(nxt, cur, k);
  }
  if (val[cur] % k && par)
    val[cur]++;
  else
    val[par]++;
}
int32_t main() {
  cin.tie(0), cin.sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long i = 1; i < n; i++) {
      cin >> u >> v;
      a[u].push_back(v);
      a[v].push_back(u);
    }
    ans[1] = 1;
    for (long long i = 1; i < n; i++) ans[1] = ans[1] * 2 % MOD;
    for (long long i = 2; i <= n; i++) {
      if ((n - 1) % i) continue;
      memset(val, 0, sizeof(val));
      dfs(1, 0, i);
      ans[i] = 1;
      for (long long j = 1; j <= n; j++)
        if (val[j] % i) ans[i] = 0;
    }
    for (long long i = n; i > 0; i--) {
      for (long long j = i + i; j <= n; j += i) ans[i] -= ans[j];
    }
    for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
    printf("\n");
    for (long long i = 1; i <= n; i++) a[i].clear(), val[i] = ans[i] = 0;
  }
  return 0;
}

```