#include <bits/stdc++.h>
using namespace std;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
const int MOD = 998244353;
int n;
vector<int> adj[MAXN];
int ans[MAXN + 1];
int par[MAXN], sz[MAXN];
void dfsinit(int at) {
  sz[at] = 1;
  for (int to : adj[at]) {
    if (to == par[at]) continue;
    par[to] = at;
    dfsinit(to);
    sz[at] += sz[to];
  }
}
void solve() {
  par[0] = -1;
  dfsinit(0);
  for (int i = (1); i <= (n); ++i) ans[i] = 0;
  ans[1] = 1;
  for (int i = (0); i < (n - 1); ++i) ans[1] = (long long)ans[1] * 2 % MOD;
  for (int i = (2); i <= (n); ++i)
    if ((n - 1) % i == 0) {
      bool ok = true;
      for (int j = (0); j < (n); ++j)
        if ((sz[j] - 1) % i != 0 && sz[j] % i != 0) ok = false;
      if (ok) ++ans[i];
    }
  for (int i = n; i >= 1; --i)
    for (int j = i + i; j <= n; j += i) ans[i] -= ans[j];
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n - 1); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  solve();
  for (int i = (1); i <= (n); ++i) {
    if (i != 1) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
  for (int i = (0); i < (n); ++i) adj[i].clear();
}
int main() {
  int ncase;
  scanf("%d", &ncase);
  for (int i = (1); i <= (ncase); ++i) run();
  return 0;
}

### 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;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
const int MOD = 998244353;
int n;
vector<int> adj[MAXN];
int ans[MAXN + 1];
int par[MAXN], sz[MAXN];
void dfsinit(int at) {
  sz[at] = 1;
  for (int to : adj[at]) {
    if (to == par[at]) continue;
    par[to] = at;
    dfsinit(to);
    sz[at] += sz[to];
  }
}
void solve() {
  par[0] = -1;
  dfsinit(0);
  for (int i = (1); i <= (n); ++i) ans[i] = 0;
  ans[1] = 1;
  for (int i = (0); i < (n - 1); ++i) ans[1] = (long long)ans[1] * 2 % MOD;
  for (int i = (2); i <= (n); ++i)
    if ((n - 1) % i == 0) {
      bool ok = true;
      for (int j = (0); j < (n); ++j)
        if ((sz[j] - 1) % i != 0 && sz[j] % i != 0) ok = false;
      if (ok) ++ans[i];
    }
  for (int i = n; i >= 1; --i)
    for (int j = i + i; j <= n; j += i) ans[i] -= ans[j];
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n - 1); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  solve();
  for (int i = (1); i <= (n); ++i) {
    if (i != 1) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
  for (int i = (0); i < (n); ++i) adj[i].clear();
}
int main() {
  int ncase;
  scanf("%d", &ncase);
  for (int i = (1); i <= (ncase); ++i) run();
  return 0;
}

```