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<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int Mod=924844033;
const int G=5;

struct Road{
	int e,next;
}r[410000];int R[210000],num;

void addr(int x,int y){
	r[++num]=(Road){y,R[x]};
	R[x]=num;
}

int sz[210000],fac[210000],ifac[210000];
int g[1048580],A[1048580],B[1048580],wn[1048580],iwn[1048580],rev[1048580];
int n,N,L;

int pow(int x,int l){
	if(l==0)return 1;
	int t=pow(x,l/2);
	t=(ll)t*t%Mod;
	if(l&1)t=(ll)t*x%Mod;
	return t;
}

void dfs(int u,int fa){
	sz[u]=1;
	for(int i=R[u];i;i=r[i].next){
		int v=r[i].e;
		if(v==fa)continue;
		dfs(v,u);
		sz[u]+=sz[v];
	}
}

void dfs_1(int u,int fa){
	g[N]++;
	if(u!=1)g[N-sz[u]]--;
	for(int i=R[u];i;i=r[i].next){
		int v=r[i].e;
		if(v==fa)continue;
		g[sz[v]]--;
		dfs_1(v,u);
	}
}

void NTT(int *a,int tag){
	for(int i=0;i<n;i++){if(rev[i]>i)swap(a[i],a[rev[i]]);}
	for(int i=1;i<n;i<<=1){
		int bot=(tag==1?wn[i<<1]:iwn[i<<1]);
		for(int j=0;j<n;j+=(i<<1)){
			for(int k=0,w=1;k<i;k++,w=(ll)w*bot%Mod){
				int x=a[j+k],y=(ll)w*a[i+j+k]%Mod;
				a[j+k]=(x+y)%Mod;a[i+j+k]=(x-y+Mod)%Mod;
			}
		}
	}if(tag==-1){
		int w=pow(n,Mod-2);
		for(int i=0;i<n;i++){
			a[i]=(ll)a[i]*w%Mod;
		}
	}
}

int main(){
	int x,y;
	fac[0]=1;for(int i=1;i<=200000;i++)fac[i]=(ll)fac[i-1]*i%Mod;
	ifac[200000]=pow(fac[200000],Mod-2);for(int i=199999;i>=0;i--)ifac[i]=(ll)ifac[i+1]*(i+1)%Mod;
	scanf("%d",&N);
	for(int i=1;i<N;i++){
		scanf("%d %d",&x,&y);
		addr(x,y);addr(y,x);
	}dfs(1,0);
	dfs_1(1,0);
	for(int i=1;i<=N;i++){
		g[i]=(g[i]+Mod)%Mod;
		A[i]=(ll)g[i]*fac[i]%Mod;
		B[i]=ifac[N-i];
	}N<<=1;
	for(n=1;n<=N;n<<=1)L++;
	for(int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
	for(int i=0;i<=L;i++){
		wn[1<<i]=pow(G,(Mod-1)>>i);
		iwn[1<<i]=pow(wn[1<<i],Mod-2);
	}NTT(A,1);NTT(B,1);
	for(int i=0;i<n;i++){
		A[i]=(ll)A[i]*B[i]%Mod;
	}NTT(A,-1);
	N>>=1;
	for(int i=N+1;i<=N<<1;i++){
		printf("%d\n",(ll)A[i]*ifac[i-N]%Mod);
	}return 0;
}