#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:150777216")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rnd(123);
template <typename T1, typename T2>
inline void chkmin(T1& x, const T2& y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1& x, const T2& y) {
  if (x < y) x = y;
}
const int MAXN = 1e5 + 228;
vector<int> g[MAXN];
int mod = 998244353;
int mul(int a, int b) { return 1ll * a * b % mod; }
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int sub(int a, int b) {
  if (a - b < 0) return a - b + mod;
  return a - b;
}
int binpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int inv(int x) { return binpow(x, mod - 2); }
int fact[MAXN];
int rfact[MAXN];
int cnk(int n, int k) {
  if (n < 0 || k > n) return 0;
  return mul(fact[n], mul(rfact[k], rfact[n - k]));
}
int k;
int dfs(int v, int p) {
  int cnt = 0;
  for (int i : g[v]) {
    if (i == p) continue;
    int x = dfs(i, v);
    if (x == -1) return -1;
    cnt += x;
  }
  if (cnt % k == 0) {
    if (p == v) return 0;
    return 1;
  }
  if ((cnt + 1) % k == 0 && v != p) return 0;
  return -1;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout.precision(20), cout.setf(ios::fixed);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) g[i].clear();
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    vector<int> dd;
    vector<int> ans(n + 1);
    ans[1] = binpow(2, n - 1);
    int x = n - 1;
    for (int i = 1; i * i <= x; ++i) {
      if (x % i == 0) {
        dd.push_back(i);
        dd.push_back(x / i);
      }
    }
    for (int i : dd) {
      if (i > 1) {
        k = i;
        ans[i] = (dfs(1, 1) == 0);
      }
    }
    for (int i = n; i >= 1; --i) {
      for (int j = i * 2; j <= n; j += i) {
        ans[i] = sub(ans[i], ans[j]);
      }
    }
    for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
    cout << '\n';
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#pragma comment(linker, "/STACK:150777216")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rnd(123);
template <typename T1, typename T2>
inline void chkmin(T1& x, const T2& y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1& x, const T2& y) {
  if (x < y) x = y;
}
const int MAXN = 1e5 + 228;
vector<int> g[MAXN];
int mod = 998244353;
int mul(int a, int b) { return 1ll * a * b % mod; }
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int sub(int a, int b) {
  if (a - b < 0) return a - b + mod;
  return a - b;
}
int binpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int inv(int x) { return binpow(x, mod - 2); }
int fact[MAXN];
int rfact[MAXN];
int cnk(int n, int k) {
  if (n < 0 || k > n) return 0;
  return mul(fact[n], mul(rfact[k], rfact[n - k]));
}
int k;
int dfs(int v, int p) {
  int cnt = 0;
  for (int i : g[v]) {
    if (i == p) continue;
    int x = dfs(i, v);
    if (x == -1) return -1;
    cnt += x;
  }
  if (cnt % k == 0) {
    if (p == v) return 0;
    return 1;
  }
  if ((cnt + 1) % k == 0 && v != p) return 0;
  return -1;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout.precision(20), cout.setf(ios::fixed);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) g[i].clear();
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    vector<int> dd;
    vector<int> ans(n + 1);
    ans[1] = binpow(2, n - 1);
    int x = n - 1;
    for (int i = 1; i * i <= x; ++i) {
      if (x % i == 0) {
        dd.push_back(i);
        dd.push_back(x / i);
      }
    }
    for (int i : dd) {
      if (i > 1) {
        k = i;
        ans[i] = (dfs(1, 1) == 0);
      }
    }
    for (int i = n; i >= 1; --i) {
      for (int j = i * 2; j <= n; j += i) {
        ans[i] = sub(ans[i], ans[j]);
      }
    }
    for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
    cout << '\n';
  }
}

```