#include <bits/stdc++.h>
using namespace std;
typedef const int& ci;
vector<int> sn[100000 + 1];
int deg[100000 + 1], que[100000 + 1], par[100000 + 1], ans[100000],
    cnt[100000 + 1];
int check(ci n, ci k) {
  memset(cnt + 1, 0, n << 2);
  for (int i = 1; i < n; ++i) {
    int u = que[i];
    if (cnt[u] % k) {
      if ((++cnt[u]) % k) return 0;
    } else
      ++cnt[par[u]];
  }
  return 1;
}
__inline__ __attribute__((always_inline)) void test() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) sn[i].clear();
  memset(deg + 1, 0, n << 2);
  for (int i = n, u, v; --i;) {
    scanf("%d%d", &u, &v);
    sn[u].push_back(v), sn[v].push_back(u), ++deg[u], ++deg[v];
  }
  int tail = 0;
  for (int i = 1; i <= n; ++i)
    if (deg[i] == 1) que[++tail] = i;
  for (int head = 1; head < n; ++head) {
    int u = que[head];
    for (auto v : sn[u])
      if (deg[v] > 1) {
        par[u] = v;
        if (--deg[v] == 1) que[++tail] = v;
      }
  }
  int m = n - 1;
  ans[1] = 1;
  for (int i = m; i; --i, ans[1] = (ans[1] << 1) % 998244353)
    ;
  for (int i = m; i > 1; --i)
    if (!(m % i)) {
      ans[i] = check(n, i);
      for (int j = i << 1; j <= m; j += i) ans[i] -= ans[j];
      ans[1] -= ans[i];
    } else
      ans[i] = 0;
  for (int i = 1; i < n; ++i) printf("%d ", ans[i]);
  puts("0");
}
int main() {
  int t;
  for (scanf("%d", &t); t; test(), --t)
    ;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
typedef const int& ci;
vector<int> sn[100000 + 1];
int deg[100000 + 1], que[100000 + 1], par[100000 + 1], ans[100000],
    cnt[100000 + 1];
int check(ci n, ci k) {
  memset(cnt + 1, 0, n << 2);
  for (int i = 1; i < n; ++i) {
    int u = que[i];
    if (cnt[u] % k) {
      if ((++cnt[u]) % k) return 0;
    } else
      ++cnt[par[u]];
  }
  return 1;
}
__inline__ __attribute__((always_inline)) void test() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) sn[i].clear();
  memset(deg + 1, 0, n << 2);
  for (int i = n, u, v; --i;) {
    scanf("%d%d", &u, &v);
    sn[u].push_back(v), sn[v].push_back(u), ++deg[u], ++deg[v];
  }
  int tail = 0;
  for (int i = 1; i <= n; ++i)
    if (deg[i] == 1) que[++tail] = i;
  for (int head = 1; head < n; ++head) {
    int u = que[head];
    for (auto v : sn[u])
      if (deg[v] > 1) {
        par[u] = v;
        if (--deg[v] == 1) que[++tail] = v;
      }
  }
  int m = n - 1;
  ans[1] = 1;
  for (int i = m; i; --i, ans[1] = (ans[1] << 1) % 998244353)
    ;
  for (int i = m; i > 1; --i)
    if (!(m % i)) {
      ans[i] = check(n, i);
      for (int j = i << 1; j <= m; j += i) ans[i] -= ans[j];
      ans[1] -= ans[i];
    } else
      ans[i] = 0;
  for (int i = 1; i < n; ++i) printf("%d ", ans[i]);
  puts("0");
}
int main() {
  int t;
  for (scanf("%d", &t); t; test(), --t)
    ;
  return 0;
}

```