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;

#define ll long long
#define pb push_back
#define bit(a , b) (((a)>>(b))&1)

const int maxb = 19;
const int N = (1 << maxb);
const int maxn = N + 20;
const int mod = 924844033;

int bpw(int a , int b)
{
	if(!b)
		return 1;
	int x = bpw(a , b / 2);

	x = 1LL * x * x % mod;
	if(b&1)
		x = 1LL * x * a % mod;

	return x;
}

void rev(int* a)
{
	for(int i = 0; i < N; i++)
	{
		int k = 0;
		for(int j = 0; j < maxb; j++)
			if(bit(i , j))
				k |= 1 << (maxb - 1 - j);

		if(i < k)
			swap(a[i] , a[k]);
	}
}

void mul(int* a , int val)
{
	rev(a);

	for(int len = 2; len <= N; len *= 2)
	{
		int wn = bpw(5 , (mod - 1) / len);
		if(val == -1)
			wn = bpw(wn , mod - 2);
		for(int i = 0; i < N; i += len)
		{
			int nxt = len / 2 , pw = 1;
			for(int j = 0; j < nxt; j++)
			{
				int x = a[i + j] , y = a[i + j + nxt];
				a[i + j] = (x + 1LL * pw * y) % mod;
				a[i + j + nxt] = (x - 1LL * pw * y % mod + mod) % mod;
				pw = 1LL * pw * wn % mod;
			}
		}
	}

	if(val == -1)
	{
		int inv = bpw(N , mod - 2);
		for(int i = 0; i < N; i++)
			a[i] = 1LL * a[i] * inv % mod;
	}
}

vector<int> adj[maxn];

int t[maxn] , sz[maxn] , n , fac[maxn] , caf[maxn] , a[maxn] , b[maxn] , c[maxn];

void dfs(int v , int p = -1)
{
	sz[v] = 1;
	for(auto u : adj[v])
		if(u != p)
		{
			dfs(u , v);
			sz[v] += sz[u];
			t[sz[u]]++;
		}

	t[n - sz[v]]++;
}

int ch(int s , int r)
{
	return 1LL * fac[s] * caf[r] % mod * caf[s - r] % mod;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	fac[0] = 1;
	for(int i = 1; i < maxn; i++)
		fac[i] = 1LL * fac[i - 1] * i % mod;
	
	caf[maxn - 1] = bpw(fac[maxn - 1] , mod - 2);
	for(int i = maxn - 1; i > 0; i--)
		caf[i - 1] = 1LL * caf[i] * i % mod;

	cin >> n;
	for(int i = 0; i < n - 1; i++)
	{
		int a , b;
		cin >> a >> b;
		a-- , b--;

		adj[a].pb(b);
		adj[b].pb(a);
	}

	dfs(0);
	for(int i = 0; i <= n; i++)
	{
		a[i] = 1LL * t[i] * fac[i] % mod;
		b[n - i] = caf[i];
	}

	mul(a , 1);
	mul(b , 1);
	for(int i = 0; i < N; i++)
		a[i] = 1LL * a[i] * b[i] % mod;
	mul(a , -1);

	for(int i = 1; i <= n; i++)
	{
		c[i] = 1LL * a[i + n] * caf[i] % mod;
		c[i] = (1LL * n * ch(n , i) - c[i] + mod) % mod;
		cout << c[i] << endl;
	}
}