#include <bits/stdc++.h>
using namespace std;
void main0();
int main() {
  clock_t start, end;
  ios::sync_with_stdio(false);
  cin.tie(0);
  main0();
  return 0;
}
const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 2e5 + 5;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double Pi = acos(-1.0);
mt19937 rnd(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long qpow(long long n, long long m) {
  long long ret = 1;
  while (m) {
    if (m & 1) ret = ret * n % mod;
    n = n * n % mod;
    m >>= 1;
  }
  return ret;
}
vector<int> e[N];
int ans[N];
int succ;
int dfs(int k, int u, int fa) {
  int cnt = 0;
  for (int v : e[u])
    if (v != fa) {
      cnt += dfs(k, v, u);
    }
  if (cnt % k == 0) return 1;
  if (cnt % k == k - 1) return 0;
  succ = 0;
  return 0;
}
void main0() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      e[i].clear();
      ans[i] = (i == 1 ? qpow(2, n - 1) : 0);
    }
    for (int i = 1; i <= n - 1; ++i) {
      int u, v;
      cin >> u >> v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    for (int i = 2; i <= n - 1; ++i)
      if ((n - 1) % i == 0) {
        succ = 1;
        dfs(i, 1, -1);
        if (succ) ans[1]--, ans[i]++;
      }
    for (int i = 2; i <= n; ++i)
      if (ans[i]) {
        for (int j = 2; i * j <= n; ++j)
          if (ans[i * j]) {
            ans[1]++, ans[i]--;
            break;
          }
      }
    ans[1] += mod;
    ans[1] %= mod;
    for (int i = 1; i <= n; ++i) {
      cout << ans[i] << ' ';
    }
    cout << endl;
  }
}

### 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;
void main0();
int main() {
  clock_t start, end;
  ios::sync_with_stdio(false);
  cin.tie(0);
  main0();
  return 0;
}
const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 2e5 + 5;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double Pi = acos(-1.0);
mt19937 rnd(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long qpow(long long n, long long m) {
  long long ret = 1;
  while (m) {
    if (m & 1) ret = ret * n % mod;
    n = n * n % mod;
    m >>= 1;
  }
  return ret;
}
vector<int> e[N];
int ans[N];
int succ;
int dfs(int k, int u, int fa) {
  int cnt = 0;
  for (int v : e[u])
    if (v != fa) {
      cnt += dfs(k, v, u);
    }
  if (cnt % k == 0) return 1;
  if (cnt % k == k - 1) return 0;
  succ = 0;
  return 0;
}
void main0() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      e[i].clear();
      ans[i] = (i == 1 ? qpow(2, n - 1) : 0);
    }
    for (int i = 1; i <= n - 1; ++i) {
      int u, v;
      cin >> u >> v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    for (int i = 2; i <= n - 1; ++i)
      if ((n - 1) % i == 0) {
        succ = 1;
        dfs(i, 1, -1);
        if (succ) ans[1]--, ans[i]++;
      }
    for (int i = 2; i <= n; ++i)
      if (ans[i]) {
        for (int j = 2; i * j <= n; ++j)
          if (ans[i * j]) {
            ans[1]++, ans[i]--;
            break;
          }
      }
    ans[1] += mod;
    ans[1] %= mod;
    for (int i = 1; i <= n; ++i) {
      cout << ans[i] << ' ';
    }
    cout << endl;
  }
}

```