#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 998244353;
int n;
vector<int> dsk[N];
long long f[N];
int cur[N];
bool ok = 1;
void dfs(int u, int pre, int val) {
  if (((int)(dsk[u]).size()) == 1 && u != 1) {
    cur[u] = 1;
    return;
  }
  for (int v : dsk[u])
    if (v != pre) {
      dfs(v, u, val);
      cur[u] += cur[v];
    }
  if (cur[u] % val == 0)
    cur[u] = 1;
  else if (u != pre && (cur[u] + 1) % val == 0)
    cur[u] = 0;
  else
    ok = 0;
}
int solve(int val) {
  for (int i = 1; i <= n; ++i) cur[i] = 0;
  ok = 1;
  dfs(1, 1, val);
  return ok;
}
void gogo() {
  cin >> n;
  for (int i = 1; i <= n; ++i) dsk[i].clear();
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    dsk[u].push_back(v);
    dsk[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i) f[i] = 0;
  f[1] = 1;
  for (int i = 1; i < n; ++i) (f[1] *= 2) %= mod;
  for (int i = 2; i <= n - 1; ++i)
    if ((n - 1) % i == 0) f[i] = solve(i);
  for (int i = n; i >= 1; --i)
    for (int j = 2 * i; j <= n; j += i) f[i] = (f[i] - f[j] + mod) % mod;
  for (int i = 1; i <= n; ++i) {
    if (i >= 2) assert(f[i] <= 1);
    cout << f[i] << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) gogo();
}

### 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 int N = 1e5 + 5;
const int mod = 998244353;
int n;
vector<int> dsk[N];
long long f[N];
int cur[N];
bool ok = 1;
void dfs(int u, int pre, int val) {
  if (((int)(dsk[u]).size()) == 1 && u != 1) {
    cur[u] = 1;
    return;
  }
  for (int v : dsk[u])
    if (v != pre) {
      dfs(v, u, val);
      cur[u] += cur[v];
    }
  if (cur[u] % val == 0)
    cur[u] = 1;
  else if (u != pre && (cur[u] + 1) % val == 0)
    cur[u] = 0;
  else
    ok = 0;
}
int solve(int val) {
  for (int i = 1; i <= n; ++i) cur[i] = 0;
  ok = 1;
  dfs(1, 1, val);
  return ok;
}
void gogo() {
  cin >> n;
  for (int i = 1; i <= n; ++i) dsk[i].clear();
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    dsk[u].push_back(v);
    dsk[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i) f[i] = 0;
  f[1] = 1;
  for (int i = 1; i < n; ++i) (f[1] *= 2) %= mod;
  for (int i = 2; i <= n - 1; ++i)
    if ((n - 1) % i == 0) f[i] = solve(i);
  for (int i = n; i >= 1; --i)
    for (int j = 2 * i; j <= n; j += i) f[i] = (f[i] - f[j] + mod) % mod;
  for (int i = 1; i <= n; ++i) {
    if (i >= 2) assert(f[i] <= 1);
    cout << f[i] << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) gogo();
}

```