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 <stdio.h>
#include <iostream>
#include <cstring>
typedef long long ll;
const int N=530000;
const int mod=924844033;
const int G=5;
int n,head[N],now=0,siz[N],A[N],B[N];
int tim[N],fac[N],ifac[N],rev[N];
using namespace std;

struct node
{
	int t,next;
}adj[N<<1];
void add(int s,int t)
{
	adj[++now]=(node){t,head[s]};
	head[s]=now;
}
int calc(int a,int b)
{
	int ans=1,tmp=a;
	while (b)
	{
		if (b&1) ans=(ll)ans*tmp%mod;
		tmp=(ll)tmp*tmp%mod,b>>=1;
	}
	return ans;
}
void dfs(int s,int fa)
{
	int i;siz[s]=1;
	for (i=head[s];i;i=adj[i].next)
	if (adj[i].t!=fa) dfs(adj[i].t,s),siz[s]+=siz[adj[i].t];
	for (i=head[s];i;i=adj[i].next)
	if (adj[i].t!=fa)
	{
		if (tim[siz[adj[i].t]]>0) tim[siz[adj[i].t]]--;
		else tim[siz[adj[i].t]]=mod-1;
	}
	if (tim[n-siz[s]]>0) tim[n-siz[s]]--;
	else tim[n-siz[s]]=mod-1;
	tim[n]=(tim[n]+1)%mod;
}
void NTT(int *a,int n,int flag)
{
	int i,j,k,x,y,w,wn;
	for (i=1;i<n;i++)
	if (i<rev[i]) swap(a[i],a[rev[i]]);
	for (i=1;i<n;i<<=1)
	{
		wn=calc(G,(mod-1)/(i<<1));
		if (flag) wn=calc(wn,mod-2);
		for (j=0;j<n;j+=(i<<1))
		for (k=0,w=1;k<i;k++,w=(ll)w*wn%mod)
		x=a[j+k],y=(ll)a[j+k+i]*w%mod,a[j+k]=(x+y)%mod,a[j+k+i]=(x-y+mod)%mod;
	}
	if (flag)
	{
		int inv=calc(n,mod-2);
		for (i=0;i<n;i++) a[i]=(ll)a[i]*inv%mod;
	}
}
int main()
{
	#ifdef Kay
		freopen ("agc005_f.in","r",stdin);
		freopen ("agc005_f.out","w",stdout);
	#endif
	scanf ("%d",&n);
	int i,s,t;
	for (i=1;i<n;i++)
	scanf ("%d %d",&s,&t),add(s,t),add(t,s);
	dfs(1,0);
	fac[0]=ifac[0]=1;
	for (i=1;i<=n;i++) fac[i]=(ll)fac[i-1]*i%mod;
	ifac[n]=calc(fac[n],mod-2);
	for (i=n-1;i>=1;i--) ifac[i]=(ll)ifac[i+1]*(i+1)%mod;
	int tot,len;
	for (tot=0,len=1;len<=n;len<<=1,tot++);len<<=1,tot++;
	for (i=1;i<=n;i++) A[n-i]=(ll)tim[i]*fac[i]%mod;
	for (i=0;i<=n;i++) B[i]=ifac[i];
	for (i=1;i<len;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(tot-1));
	NTT(A,len,0),NTT(B,len,0);
	for (i=0;i<len;i++) A[i]=(ll)A[i]*B[i]%mod;
	NTT(A,len,1);
	for (i=1;i<=n;i++) printf ("%d\n",(ll)A[n-i]*ifac[i]%mod);
	return 0;
}
