#include <bits/stdc++.h>
using namespace std;
vector<long long> ch[100005];
long long num[100005], B = 998244353;
long long dfs(long long now, long long from, long long mod) {
  long long a, b, c;
  c = 0;
  for (auto a : ch[now]) {
    if (a == now || a == from) continue;
    b = dfs(a, now, mod);
    if (b == -1) return -1;
    c += b;
  }
  if ((ch[now].size() - c) % mod == 0) return 1;
  if (ch[now].size() - c > 0 && (ch[now].size() - c - 1) % mod == 0 &&
      from != 0)
    return 0;
  return -1;
}
void solve() {
  long long i, j, k, a, b, c, n;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) num[i] = 0;
  for (i = 1; i <= n; i++) {
    ch[i].clear();
  }
  for (i = 1; i < n; i++) {
    scanf("%lld%lld", &a, &b);
    ch[a].push_back(b);
    ch[b].push_back(a);
  }
  for (i = 2; i <= n; i++) {
    if ((2 * n - 2) % i) continue;
    if (dfs(1, 0, i) >= 0) num[i] = 1;
  }
  num[1] = 1;
  for (i = 1; i < n; i++) num[1] = (num[1] * 2) % B;
  c = 0;
  for (i = n; i > 1; i--) {
    for (j = i * 2; j <= n; j += i) {
      if (num[j] == 1) num[i] = 0;
    }
    c += num[i];
  }
  num[1] -= c;
  while (num[1] < 0) num[1] += B;
  for (i = 1; i <= n; i++) printf("%lld ", num[i]);
  printf("\n");
}
int main() {
  int T = 1;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
}

### 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;
vector<long long> ch[100005];
long long num[100005], B = 998244353;
long long dfs(long long now, long long from, long long mod) {
  long long a, b, c;
  c = 0;
  for (auto a : ch[now]) {
    if (a == now || a == from) continue;
    b = dfs(a, now, mod);
    if (b == -1) return -1;
    c += b;
  }
  if ((ch[now].size() - c) % mod == 0) return 1;
  if (ch[now].size() - c > 0 && (ch[now].size() - c - 1) % mod == 0 &&
      from != 0)
    return 0;
  return -1;
}
void solve() {
  long long i, j, k, a, b, c, n;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) num[i] = 0;
  for (i = 1; i <= n; i++) {
    ch[i].clear();
  }
  for (i = 1; i < n; i++) {
    scanf("%lld%lld", &a, &b);
    ch[a].push_back(b);
    ch[b].push_back(a);
  }
  for (i = 2; i <= n; i++) {
    if ((2 * n - 2) % i) continue;
    if (dfs(1, 0, i) >= 0) num[i] = 1;
  }
  num[1] = 1;
  for (i = 1; i < n; i++) num[1] = (num[1] * 2) % B;
  c = 0;
  for (i = n; i > 1; i--) {
    for (j = i * 2; j <= n; j += i) {
      if (num[j] == 1) num[i] = 0;
    }
    c += num[i];
  }
  num[1] -= c;
  while (num[1] < 0) num[1] += B;
  for (i = 1; i <= n; i++) printf("%lld ", num[i]);
  printf("\n");
}
int main() {
  int T = 1;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
}

```