#include <bits/stdc++.h>
using namespace std;
long long PosMod(long long a) { return a >= 0 ? a : a + 998244353; }
long long AddMod(long long a, long long b) {
  return PosMod((a + b) % 998244353);
}
long long MultMod(long long a, long long b) {
  return PosMod((a * b) % 998244353);
}
long long powll(long long a, long long n) {
  long long r = 1;
  for (long long b = n; b > 0; b >>= 1, a = MultMod(a, a))
    if (b & 1) r = MultMod(r, a);
  return r;
}
vector<vector<int>> adj;
bool fail;
int dfs(int here, int p, int targetK) {
  if (fail) return 0;
  int incoming = 0;
  for (int there : adj[here]) {
    if (there == p) continue;
    incoming += dfs(there, here, targetK);
  }
  if (incoming % targetK == 0)
    return 1;
  else if ((incoming + 1) % targetK == 0 && here != 1)
    return 0;
  else {
    fail = true;
    return 0;
  }
}
int n;
void solve() {
  cin >> n;
  adj = vector<vector<int>>(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  vector<long long> f = vector<long long>(n + 1);
  f[1] = powll(2, (long long)n - 1);
  for (int k = 2; k <= n - 1; k++) {
    if ((n - 1) % k != 0) continue;
    fail = false;
    dfs(1, 0, k);
    f[k] = fail ? 0 : 1;
  }
  vector<long long> h = vector<long long>(n + 1);
  h[1] = f[1];
  for (int i = 2; i <= n; i++) {
    long long val = f[i];
    if (val > 0) {
      for (int k = 2; i * k <= n; k++) {
        if (f[i * k]) val = 0;
      }
    }
    h[i] = val;
    h[1] -= val;
  }
  for (int i = 1; i <= n; i++) {
    printf("%lld ", h[i]);
  }
  printf("\n");
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### 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;
long long PosMod(long long a) { return a >= 0 ? a : a + 998244353; }
long long AddMod(long long a, long long b) {
  return PosMod((a + b) % 998244353);
}
long long MultMod(long long a, long long b) {
  return PosMod((a * b) % 998244353);
}
long long powll(long long a, long long n) {
  long long r = 1;
  for (long long b = n; b > 0; b >>= 1, a = MultMod(a, a))
    if (b & 1) r = MultMod(r, a);
  return r;
}
vector<vector<int>> adj;
bool fail;
int dfs(int here, int p, int targetK) {
  if (fail) return 0;
  int incoming = 0;
  for (int there : adj[here]) {
    if (there == p) continue;
    incoming += dfs(there, here, targetK);
  }
  if (incoming % targetK == 0)
    return 1;
  else if ((incoming + 1) % targetK == 0 && here != 1)
    return 0;
  else {
    fail = true;
    return 0;
  }
}
int n;
void solve() {
  cin >> n;
  adj = vector<vector<int>>(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  vector<long long> f = vector<long long>(n + 1);
  f[1] = powll(2, (long long)n - 1);
  for (int k = 2; k <= n - 1; k++) {
    if ((n - 1) % k != 0) continue;
    fail = false;
    dfs(1, 0, k);
    f[k] = fail ? 0 : 1;
  }
  vector<long long> h = vector<long long>(n + 1);
  h[1] = f[1];
  for (int i = 2; i <= n; i++) {
    long long val = f[i];
    if (val > 0) {
      for (int k = 2; i * k <= n; k++) {
        if (f[i * k]) val = 0;
      }
    }
    h[i] = val;
    h[1] -= val;
  }
  for (int i = 1; i <= n; i++) {
    printf("%lld ", h[i]);
  }
  printf("\n");
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```