Problem: One day, Takahashi was given the following problem from Aoki:

* You are given a tree with N vertices and an integer K. The vertices are numbered 1 through N. The edges are represented by pairs of integers (a_i, b_i).
* For a set S of vertices in the tree, let f(S) be the minimum number of the vertices in a subtree of the given tree that contains all vertices in S.
* There are <image> ways to choose K vertices from the trees. For each of them, let S be the set of the chosen vertices, and find the sum of f(S) over all <image> ways.
* Since the answer may be extremely large, print it modulo 924844033(prime).



Since it was too easy for him, he decided to solve this problem for all K = 1,2,...,N.

Constraints

* 2 ≦ N ≦ 200,000
* 1 ≦ a_i, b_i ≦ N
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}


Output

Print N lines. The i-th line should contain the answer to the problem where K=i, modulo 924844033.

Examples

Input

3
1 2
2 3


Output

3
7
3


Input

4
1 2
1 3
1 4


Output

4
15
13
4


Input

7
1 2
2 3
2 4
4 5
4 6
6 7


Output

7
67
150
179
122
45
7

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int p = 924844033;
int pow(int a, int n)
{
	int res = 1;
	for (; n; n >>= 1, a = (ll)a * a % p) {
		if (n & 1) {
			res = (ll)res * a % p;
		}
	}
	return res;
}
void ntt(vector<int> &a, bool x = false)
{
	int n = a.size();
	vector<int> w(n), rev(n);
	w[0] = 1;
	w[1] = pow(5, (p - 1) / n);
	if (x) {
		w[1] = pow(w[1], p - 2);
	}
	for (int i = 2; i < n; i++) {
		w[i] = (ll)w[i - 1] * w[1] % p;
	}
	for (int i = 1, j = n / 2; i + 1 < n; i++) {
		if (i < j) {
			swap(a[i], a[j]);
		}
		int k = n / 2;
		while (j >= k) {
			j -= k;
			k /= 2;
		}
		j += k;
	}
	for (int l = 1; l < n; l <<= 1) {
		for (int i = 0; i < n; i += l << 1) {
			int t = 0;
			for (int j = i; j < i + l; j++) {
				int u = a[j], v = (ll)a[j + l] * w[t] % p;
				a[j] = (u + v) % p;
				a[j + l] = (u - v + p) % p;
				t += n / l / 2;
			}
		}
	}
	if (x) {
		int t = pow(n, p - 2);
		for (int i = 0; i < n; i++) {
			a[i] = (ll)a[i] * t % p;
		}
	}
}
void mul(vector<int> &a, vector<int> b)
{
	int n = a.size(), m = 1;
	while (m >> 1 < n) {
		m <<= 1;
	}
	vector<int> A(m, 0), B(m, 0);
	copy(a.begin(), a.end(), A.begin());
	copy(b.begin(), b.end(), B.begin());
	ntt(A);
	ntt(B);
	for (int i = 0; i < m; i++) {
		A[i] = (ll)A[i] * B[i] % p;
	}
	ntt(A, true);
	a = A;
}
vector<int> adj[200005];
int sz[200005];
void init(int u, int fa = -1)
{
	sz[u] = 1;
	for (int v : adj[u]) {
		if (v == fa) {
			continue;
		}
		init(v, u);
		sz[u] += sz[v];
	}
}
int main()
{
	int n;
	cin >> n;
	for (int i = 1; i < n; i++) {
		int u, v;
		cin >> u >> v;
		adj[--u].push_back(--v);
		adj[v].push_back(u);
	}
	init(0);
	vector<int> val(n + 1, 0), fact(n + 1), inv(n + 1), rev(n + 1), ans(n + 1);
	val[n] = n;
	for (int i = 1; i < n; i++) {
		val[sz[i]]--;
		val[n - sz[i]]--;
	}
	fact[0] = inv[0] = 1;
	for (int i = 1; i <= n; i++) {
		fact[i] = (ll)fact[i - 1] * i % p;
		inv[i] = pow(fact[i], p - 2);
	}
	for (int i = 0; i <= n; i++) {
		val[i] = (ll)(val[i] + p) * fact[i] % p;
		rev[i] = inv[n - i];
	}
	mul(val, rev);
	for (int i = 1; i <= n; i++) {
		ans[i] = (ll)val[i + n] * inv[i] % p;
	}
	copy(ans.begin() + 1, ans.end(), ostream_iterator<int>(cout, "\n"));
	return 0;
}