#include <bits/stdc++.h>
using namespace std;
void dbg() { std::cout << "  #\n"; }
template <typename T, typename... Args>
void dbg(T a, Args... args) {
  std::cout << a << ' ';
  dbg(args...);
}
const long long maxn = 2e5 + 10;
const long long MAX = 1000;
const long long inf = 0x3f3f3f3f;
const long long mod = 998244353;
long long n, m;
long long a[maxn];
long long f[maxn];
long long ans[maxn];
vector<long long> A[maxn];
long long getint() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  long long x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x;
}
bool isok;
void dfs(long long now, long long fa, long long k) {
  for (auto to : A[now])
    if (to != fa) {
      dfs(to, now, k);
    }
  if (a[now] % k == 0) {
    a[fa]++;
  } else {
    if (fa != 0) {
      a[now]++;
      if (a[now] % k != 0) {
        isok = false;
      }
    }
  }
}
void solve() {
  n = getint();
  for (long long i = (1); i <= (n); ++i) ans[i] = 0, A[i].clear();
  for (long long i = (1); i <= (n - 1); ++i) {
    long long x, y;
    x = getint();
    y = getint();
    A[x].push_back(y);
    A[y].push_back(x);
  }
  ans[1] = 1;
  for (long long i = (1); i <= (n - 1); ++i) ans[1] = (ans[1] * 2ll % mod);
  for (long long i = (2); i <= (n); ++i)
    if ((n - 1) % i == 0) {
      for (long long j = (1); j <= (n); ++j) a[j] = 0;
      isok = 1;
      dfs(1, 0, i);
      if (isok) ans[i]++;
    }
  for (long long i = (n); i >= (1); --i) {
    long long j = i + i;
    while (j <= n) {
      ans[i] -= ans[j];
      ans[i] = (ans[i] + mod) % mod;
      j = j + i;
    }
  }
  for (long long i = (1); i <= (n); ++i) printf("%d ", ans[i] % mod);
  printf("\n");
}
signed main() {
  long long TestCase = 1;
  cin >> TestCase;
  while (TestCase--) {
    solve();
  }
  char EndFile = getchar();
  EndFile = getchar();
  return 0;
}

### Prompt

Generate 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;
void dbg() { std::cout << "  #\n"; }
template <typename T, typename... Args>
void dbg(T a, Args... args) {
  std::cout << a << ' ';
  dbg(args...);
}
const long long maxn = 2e5 + 10;
const long long MAX = 1000;
const long long inf = 0x3f3f3f3f;
const long long mod = 998244353;
long long n, m;
long long a[maxn];
long long f[maxn];
long long ans[maxn];
vector<long long> A[maxn];
long long getint() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  long long x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x;
}
bool isok;
void dfs(long long now, long long fa, long long k) {
  for (auto to : A[now])
    if (to != fa) {
      dfs(to, now, k);
    }
  if (a[now] % k == 0) {
    a[fa]++;
  } else {
    if (fa != 0) {
      a[now]++;
      if (a[now] % k != 0) {
        isok = false;
      }
    }
  }
}
void solve() {
  n = getint();
  for (long long i = (1); i <= (n); ++i) ans[i] = 0, A[i].clear();
  for (long long i = (1); i <= (n - 1); ++i) {
    long long x, y;
    x = getint();
    y = getint();
    A[x].push_back(y);
    A[y].push_back(x);
  }
  ans[1] = 1;
  for (long long i = (1); i <= (n - 1); ++i) ans[1] = (ans[1] * 2ll % mod);
  for (long long i = (2); i <= (n); ++i)
    if ((n - 1) % i == 0) {
      for (long long j = (1); j <= (n); ++j) a[j] = 0;
      isok = 1;
      dfs(1, 0, i);
      if (isok) ans[i]++;
    }
  for (long long i = (n); i >= (1); --i) {
    long long j = i + i;
    while (j <= n) {
      ans[i] -= ans[j];
      ans[i] = (ans[i] + mod) % mod;
      j = j + i;
    }
  }
  for (long long i = (1); i <= (n); ++i) printf("%d ", ans[i] % mod);
  printf("\n");
}
signed main() {
  long long TestCase = 1;
  cin >> TestCase;
  while (TestCase--) {
    solve();
  }
  char EndFile = getchar();
  EndFile = getchar();
  return 0;
}

```