#include <bits/stdc++.h>
using namespace std;
inline void Read(int &first) {
  int f = 1;
  first = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    first = (first << 3) + (first << 1) + c - '0', c = getchar();
  }
  first *= f;
}
const int N = 100010, p = 998244353;
int T, n, f[N];
struct Edge {
  int to, nxt;
} e[N << 1];
int hd[N], cnt;
inline void ade(int u, int v) {
  e[++cnt].to = v, e[cnt].nxt = hd[u], hd[u] = cnt;
}
int ff;
int DFS(int u, int fa, int m) {
  if (!ff) return 0;
  int tot = 0;
  for (register int i = hd[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v != fa) tot += DFS(v, u, m);
  }
  if (tot % m == 0) return 1;
  if ((tot + 1) % m == 0) return 0;
  ff = 0;
  return 0;
}
int main() {
  Read(T);
  while (T--) {
    memset(f, 0, sizeof f);
    Read(n), memset(hd, 0, sizeof hd), cnt = 0, f[1] = 1;
    for (register int i = 1, u, v; i < n; i++) {
      Read(u), Read(v), ade(u, v), ade(v, u), f[1] = 2ll * f[1] % p;
    }
    for (register int i = 2; i < n; i++) {
      if ((n - 1) % i == 0) {
        ff = 1, f[i] = DFS(1, 0, i) & ff;
      }
    }
    for (register int i = n - 1; i; i--) {
      for (register int j = i + i; j < n; j += i) f[i] -= f[j];
      f[i] = (f[i] % p + p) % p;
    }
    for (register int i = 1; i <= n; i++) printf("%d ", f[i]);
    puts("");
  }
  return 0;
}

### 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;
inline void Read(int &first) {
  int f = 1;
  first = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    first = (first << 3) + (first << 1) + c - '0', c = getchar();
  }
  first *= f;
}
const int N = 100010, p = 998244353;
int T, n, f[N];
struct Edge {
  int to, nxt;
} e[N << 1];
int hd[N], cnt;
inline void ade(int u, int v) {
  e[++cnt].to = v, e[cnt].nxt = hd[u], hd[u] = cnt;
}
int ff;
int DFS(int u, int fa, int m) {
  if (!ff) return 0;
  int tot = 0;
  for (register int i = hd[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v != fa) tot += DFS(v, u, m);
  }
  if (tot % m == 0) return 1;
  if ((tot + 1) % m == 0) return 0;
  ff = 0;
  return 0;
}
int main() {
  Read(T);
  while (T--) {
    memset(f, 0, sizeof f);
    Read(n), memset(hd, 0, sizeof hd), cnt = 0, f[1] = 1;
    for (register int i = 1, u, v; i < n; i++) {
      Read(u), Read(v), ade(u, v), ade(v, u), f[1] = 2ll * f[1] % p;
    }
    for (register int i = 2; i < n; i++) {
      if ((n - 1) % i == 0) {
        ff = 1, f[i] = DFS(1, 0, i) & ff;
      }
    }
    for (register int i = n - 1; i; i--) {
      for (register int j = i + i; j < n; j += i) f[i] -= f[j];
      f[i] = (f[i] % p + p) % p;
    }
    for (register int i = 1; i <= n; i++) printf("%d ", f[i]);
    puts("");
  }
  return 0;
}

```