#include <bits/stdc++.h>
using namespace std;
int recur_depth = 0;
const long long sz = 1e5 + 10, mod = 998244353;
vector<long long> g[sz];
long long ans[sz], dp[sz];
void dfs(long long u, long long p, long long k) {
  for (long long& v : g[u]) {
    if (v == p) continue;
    dfs(v, u, k);
    if (dp[v] % k != 0)
      dp[v]++;
    else
      dp[u]++;
  }
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    scanf("%lld", &n);
    for (long long i = 1; i <= n - 1; ++i) {
      long long u, v;
      scanf("%lld", &u), scanf("%lld", &v);
      g[u].push_back(v);
      g[v].push_back(u);
    }
    long long m = n - 1;
    ans[1] = 1;
    for (long long i = 1; i <= m; ++i) ans[1] = (ans[1] * 2) % mod;
    for (long long i = 2; i <= m; i++) {
      if (m % i == 0) {
        for (long long j = 1; j <= n; ++j) dp[j] = 0;
        dfs(1, -1, i);
        bool ok = 1;
        for (long long j = 1; j <= n; ++j) ok &= (dp[j] % i == 0);
        ans[i] = ok;
      }
    }
    for (long long i = n; i >= 1; --i)
      for (long long j = i + i; j <= n; j += i) ans[i] -= ans[j];
    for (long long i = 1; i <= n; ++i) {
      if (ans[i] < 0) ans[i] += mod;
      printf("%lld ", ans[i]);
    }
    printf("\n");
    for (long long i = 1; i <= n; ++i) g[i].clear(), ans[i] = 0;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int recur_depth = 0;
const long long sz = 1e5 + 10, mod = 998244353;
vector<long long> g[sz];
long long ans[sz], dp[sz];
void dfs(long long u, long long p, long long k) {
  for (long long& v : g[u]) {
    if (v == p) continue;
    dfs(v, u, k);
    if (dp[v] % k != 0)
      dp[v]++;
    else
      dp[u]++;
  }
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    scanf("%lld", &n);
    for (long long i = 1; i <= n - 1; ++i) {
      long long u, v;
      scanf("%lld", &u), scanf("%lld", &v);
      g[u].push_back(v);
      g[v].push_back(u);
    }
    long long m = n - 1;
    ans[1] = 1;
    for (long long i = 1; i <= m; ++i) ans[1] = (ans[1] * 2) % mod;
    for (long long i = 2; i <= m; i++) {
      if (m % i == 0) {
        for (long long j = 1; j <= n; ++j) dp[j] = 0;
        dfs(1, -1, i);
        bool ok = 1;
        for (long long j = 1; j <= n; ++j) ok &= (dp[j] % i == 0);
        ans[i] = ok;
      }
    }
    for (long long i = n; i >= 1; --i)
      for (long long j = i + i; j <= n; j += i) ans[i] -= ans[j];
    for (long long i = 1; i <= n; ++i) {
      if (ans[i] < 0) ans[i] += mod;
      printf("%lld ", ans[i]);
    }
    printf("\n");
    for (long long i = 1; i <= n; ++i) g[i].clear(), ans[i] = 0;
  }
  return 0;
}

```