#include <bits/stdc++.h>
using namespace std;
const int N = 300000, mod = 998244353;
int n;
struct side {
  int y, next;
} e[N * 2 + 9];
int lin[N + 9], cs;
void Ins(int x, int y) {
  e[++cs].y = y;
  e[cs].next = lin[x];
  lin[x] = cs;
}
void Ins2(int x, int y) {
  Ins(x, y);
  Ins(y, x);
}
void into() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) lin[i] = 0;
  cs = 0;
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    Ins2(x, y);
  }
}
int cnt[N + 9];
void Dfs_cnt(int k, int fa, int p) {
  cnt[k] = 0;
  for (int i = lin[k]; i; i = e[i].next) {
    int y = e[i].y;
    if (y == fa) continue;
    Dfs_cnt(y, k, p);
    cnt[y] % p ? ++cnt[y] : ++cnt[k];
  }
}
int ans[N + 9];
void Get_ans() {
  for (int i = 1; i <= n; ++i) ans[i] = 0;
  ans[1] = 1;
  for (int i = 1; i < n; ++i) ans[1] = 1LL * ans[1] * 2 % mod;
  for (int i = 2; i < n; ++i) {
    if ((n - 1) % i) continue;
    Dfs_cnt(1, 0, i);
    ans[i] = 1;
    for (int j = 1; j <= n; ++j)
      if (cnt[j] % i) ans[i] = 0;
  }
  for (int i = n; i >= 1; --i)
    for (int j = i << 1; j <= n; j += i) ans[i] = (ans[i] + mod - ans[j]) % mod;
}
void work() { Get_ans(); }
void outo() {
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  puts("");
}
int main() {
  int T = 1;
  scanf("%d", &T);
  for (; T--;) {
    into();
    work();
    outo();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 = 300000, mod = 998244353;
int n;
struct side {
  int y, next;
} e[N * 2 + 9];
int lin[N + 9], cs;
void Ins(int x, int y) {
  e[++cs].y = y;
  e[cs].next = lin[x];
  lin[x] = cs;
}
void Ins2(int x, int y) {
  Ins(x, y);
  Ins(y, x);
}
void into() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) lin[i] = 0;
  cs = 0;
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    Ins2(x, y);
  }
}
int cnt[N + 9];
void Dfs_cnt(int k, int fa, int p) {
  cnt[k] = 0;
  for (int i = lin[k]; i; i = e[i].next) {
    int y = e[i].y;
    if (y == fa) continue;
    Dfs_cnt(y, k, p);
    cnt[y] % p ? ++cnt[y] : ++cnt[k];
  }
}
int ans[N + 9];
void Get_ans() {
  for (int i = 1; i <= n; ++i) ans[i] = 0;
  ans[1] = 1;
  for (int i = 1; i < n; ++i) ans[1] = 1LL * ans[1] * 2 % mod;
  for (int i = 2; i < n; ++i) {
    if ((n - 1) % i) continue;
    Dfs_cnt(1, 0, i);
    ans[i] = 1;
    for (int j = 1; j <= n; ++j)
      if (cnt[j] % i) ans[i] = 0;
  }
  for (int i = n; i >= 1; --i)
    for (int j = i << 1; j <= n; j += i) ans[i] = (ans[i] + mod - ans[j]) % mod;
}
void work() { Get_ans(); }
void outo() {
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  puts("");
}
int main() {
  int T = 1;
  scanf("%d", &T);
  for (; T--;) {
    into();
    work();
    outo();
  }
  return 0;
}

```