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
#include <bits/stdc++.h>
using namespace std;

#define MAXN 1048577
#define mod 924844033

int n,fac[MAXN],inv[MAXN],size[MAXN],a[MAXN],b[MAXN],rev[MAXN],t[MAXN],l;
vector<int> edge[MAXN];

int qpow(int a,int b)
{
	int res=1;
	while(b)
	{
		if(b&1)res=(long long)res*a%mod;
		a=(long long)a*a%mod;
		b>>=1;
	}
	return res;
}

void dfs(int u,int fa)
{
	size[u]=1;
	for(auto v : edge[u])
	{
		if(v==fa)continue;
		dfs(v,u);
		size[u]+=size[v];
	}
	if(u^1)
	{
		++a[size[u]];
		++a[n-size[u]];
	}
}

void init()
{
	fac[0]=1;
	for(int i=1;i<=n;++i)fac[i]=(long long)fac[i-1]*i%mod;
	dfs(1,0);
	for(int i=1;i<n;++i)a[i]=(long long)a[i]*fac[i]%mod;
	for(int i=0;i<=n;++i)inv[i]=qpow(fac[i],mod-2);
	for(int i=0;i<=n;++i)b[n-i]=inv[i];
}

void dft(int *a)
{
	for(int i=0;i<l;++i)if(rev[i]<i)swap(a[i],a[rev[i]]);
	for(int i=1;i<l;i<<=1)
		for(int j=0;j<l;j+=(i<<1))
			for(int k=0;k<i;++k)
			{
				int w=(long long)a[i+j+k]*t[i+k]%mod;
				a[i+j+k]=a[j+k]-w;
				if(a[i+j+k]<0)a[i+j+k]+=mod;
				a[j+k]+=w;
				if(a[j+k]>=mod)a[j+k]-=mod;
			}
}

void idft(int *a)
{
	dft(a);
	int w=qpow(l,mod-2);
	for(int i=1;i<(l>>1);++i)swap(a[i],a[l-i]);
	for(int i=0;i<l;++i)a[i]=(long long)a[i]*w%mod;
}

void solve()
{
	l=1;
	while(l<=2*n)l<<=1;
	for(int i=0;i<l;++i)rev[i]=(rev[i>>1]>>1)|((i&1)?(l>>1):0);
	for(int i=1;i<l;i<<=1)
		for(int j=0,nw=1,w=qpow(5,(mod-1)/(i<<1));j<i;++j,nw=(long long)nw*w%mod)t[i+j]=nw;
	dft(a);
	dft(b);
	for(int i=0;i<l;++i)a[i]=(long long)a[i]*b[i]%mod;
	idft(a);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;++i)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	init();
	solve();
	for(int i=1;i<=n;++i)printf("%d\n",((long long)n*fac[n]%mod*inv[i]%mod*inv[n-i]%mod-(long long)inv[i]*a[n+i]%mod+mod)%mod);
}