#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 50;
const long long mod = 998244353;
vector<long long> g[N];
long long F[N];
long long H[N];
long long binpow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) {
      ret = (ret * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
void reset(long long n) {
  for (long long i = 1; i <= n; i++) {
    g[i].clear();
    F[i] = 0;
  }
}
long long getF(long long node, long long k, long long par = -1) {
  long long val = 0;
  for (auto to : g[node]) {
    if (to == par) continue;
    long long res = getF(to, k, node);
    if (res == -1) {
      return -1;
    }
    if (res == 0) val++;
  }
  if (val % k == 0) {
    return 0;
  } else if (val % k == k - 1) {
    return 1;
  } else {
    return -1;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    F[1] = binpow(2, n - 1);
    for (long long i = 2; i <= n; i++) {
      if ((n - 1) % i == 0) {
        if (getF(1, i) != -1) F[i] = 1;
      }
    }
    for (long long i = n; i >= 1; i--) {
      H[i] = F[i];
      for (long long j = i * 2; j <= n; j += i) {
        H[i] -= H[j];
      }
    }
    for (long long i = 1; i <= n; i++) cout << H[i] << ' ';
    cout << '\n';
    reset(n);
  }
}

### Prompt

Your task is to create a cpp solution 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;
const long long N = 1e5 + 50;
const long long mod = 998244353;
vector<long long> g[N];
long long F[N];
long long H[N];
long long binpow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) {
      ret = (ret * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
void reset(long long n) {
  for (long long i = 1; i <= n; i++) {
    g[i].clear();
    F[i] = 0;
  }
}
long long getF(long long node, long long k, long long par = -1) {
  long long val = 0;
  for (auto to : g[node]) {
    if (to == par) continue;
    long long res = getF(to, k, node);
    if (res == -1) {
      return -1;
    }
    if (res == 0) val++;
  }
  if (val % k == 0) {
    return 0;
  } else if (val % k == k - 1) {
    return 1;
  } else {
    return -1;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    F[1] = binpow(2, n - 1);
    for (long long i = 2; i <= n; i++) {
      if ((n - 1) % i == 0) {
        if (getF(1, i) != -1) F[i] = 1;
      }
    }
    for (long long i = n; i >= 1; i--) {
      H[i] = F[i];
      for (long long j = i * 2; j <= n; j += i) {
        H[i] -= H[j];
      }
    }
    for (long long i = 1; i <= n; i++) cout << H[i] << ' ';
    cout << '\n';
    reset(n);
  }
}

```