#include <bits/stdc++.h>
using namespace std;
const int M = 998244353;
int Tc, n, ans[100004], dp[100004];
vector<int> g[100004];
bool ok;
void dfs(int x, int par, int C) {
  if (!ok) return;
  for (auto y : g[x])
    if (y != par) dfs(y, x, C);
  if (dp[x] % C != 0) {
    if (par) dp[x]++;
    if (dp[x] % C != 0) {
      ok = 0;
      return;
    }
  } else
    dp[par]++;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) g[i].clear();
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) ans[i] = 0;
  ans[1] = 1;
  for (int i = 1; i < n; i++) ans[1] = (ans[1] + ans[1]) % M;
  for (int i = 2; i <= n - 1; i++)
    if ((n - 1) % i == 0) {
      for (int j = 0; j <= n; j++) dp[j] = 0;
      ok = 1, dfs(1, 0, i), ans[i] = ok;
    }
  for (int i = n; i; i--)
    for (int j = i + i; j <= n; j += i) ans[i] = (ans[i] - ans[j] + M) % M;
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  for (scanf("%d", &Tc); Tc--;) solve();
}

### Prompt

Your challenge is to write 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 M = 998244353;
int Tc, n, ans[100004], dp[100004];
vector<int> g[100004];
bool ok;
void dfs(int x, int par, int C) {
  if (!ok) return;
  for (auto y : g[x])
    if (y != par) dfs(y, x, C);
  if (dp[x] % C != 0) {
    if (par) dp[x]++;
    if (dp[x] % C != 0) {
      ok = 0;
      return;
    }
  } else
    dp[par]++;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) g[i].clear();
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) ans[i] = 0;
  ans[1] = 1;
  for (int i = 1; i < n; i++) ans[1] = (ans[1] + ans[1]) % M;
  for (int i = 2; i <= n - 1; i++)
    if ((n - 1) % i == 0) {
      for (int j = 0; j <= n; j++) dp[j] = 0;
      ok = 1, dfs(1, 0, i), ans[i] = ok;
    }
  for (int i = n; i; i--)
    for (int j = i + i; j <= n; j += i) ans[i] = (ans[i] - ans[j] + M) % M;
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  for (scanf("%d", &Tc); Tc--;) solve();
}

```