#include <bits/stdc++.h>
using namespace std;
const int N = 200020, MOD = 998244353;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
int T, n, u, v, f[N];
bool flag;
vector<int> e[N];
int dfs(int k, int fa, int gr) {
  int cnt = 0;
  for (int i = 0; i < e[k].size(); i++) {
    int to = e[k][i];
    if (to == fa) continue;
    cnt = cnt + dfs(to, k, gr);
  }
  if (cnt % gr == 0) return 1;
  if ((cnt + 1) % gr == 0) return 0;
  flag = 0;
  return 0;
}
int main() {
  T = read();
  while (T--) {
    n = read();
    for (int i = 1; i <= n; i++) e[i].clear();
    memset(f, 0, sizeof(f)), f[1] = 1;
    for (int i = 1; i <= n - 1; i++) {
      f[1] = f[1] * 2 % MOD;
      u = read(), v = read();
      e[u].push_back(v), e[v].push_back(u);
    }
    for (int i = 2; i <= n - 1; i++)
      if ((n - 1) % i == 0) {
        flag = 1;
        if (dfs(1, 0, i) == 1 && flag) f[i] = 1;
      }
    for (int i = n - 1; i >= 1; i--)
      for (int j = 2; i * j <= n - 1; j++) f[i] = f[i] - f[i * j];
    f[1] = (f[1] + MOD) % MOD;
    for (int i = 1; i <= n; i++) printf("%d ", f[i]);
    printf("\n");
  }
  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;
const int N = 200020, MOD = 998244353;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
int T, n, u, v, f[N];
bool flag;
vector<int> e[N];
int dfs(int k, int fa, int gr) {
  int cnt = 0;
  for (int i = 0; i < e[k].size(); i++) {
    int to = e[k][i];
    if (to == fa) continue;
    cnt = cnt + dfs(to, k, gr);
  }
  if (cnt % gr == 0) return 1;
  if ((cnt + 1) % gr == 0) return 0;
  flag = 0;
  return 0;
}
int main() {
  T = read();
  while (T--) {
    n = read();
    for (int i = 1; i <= n; i++) e[i].clear();
    memset(f, 0, sizeof(f)), f[1] = 1;
    for (int i = 1; i <= n - 1; i++) {
      f[1] = f[1] * 2 % MOD;
      u = read(), v = read();
      e[u].push_back(v), e[v].push_back(u);
    }
    for (int i = 2; i <= n - 1; i++)
      if ((n - 1) % i == 0) {
        flag = 1;
        if (dfs(1, 0, i) == 1 && flag) f[i] = 1;
      }
    for (int i = n - 1; i >= 1; i--)
      for (int j = 2; i * j <= n - 1; j++) f[i] = f[i] - f[i * j];
    f[1] = (f[1] + MOD) % MOD;
    for (int i = 1; i <= n; i++) printf("%d ", f[i]);
    printf("\n");
  }
  return 0;
}

```