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<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<string>
#include<iostream>
#include<cmath>
#include<queue>
#include<bitset>
#define maxn 800010
#define ms(x,y) memset(x,y,sizeof x)
#define inf 0x3f3f3f3f
#define fir first
#define sec second
#define lb(x) (x&(-x))
const int mod=924844033;
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
ll fastpow(ll num,ll mi){
	ll res=1;
	while(mi){
		if(mi&1)res=res*num%mod;
		num=num*num%mod;
		mi>>=1;
	}
	return res;
}
namespace ntt{
	int n,m,r[800010];
	ll w[20],g=5;
	void init(int len){
		for(n=1,m=0;n<=len;n<<=1,m++);
		for(int i=1;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)<<m-1);
	}
	void initw(int f){
		w[0]=fastpow(g,(mod-1)/n);
		if(f<0)w[0]=fastpow(w[0],mod-2);
		for(int i=1;i<m;i++)w[i]=w[i-1]*w[i-1]%mod;
	}
	void ntt(ll *a,int f){
		for(int i=1;i<n;i++)if(i<r[i])swap(a[i],a[r[i]]);
		for(int i=1,x=m-1;i<n;i<<=1,x--){
			for(int j=0;j<n;j+=i<<1){
				ll wn=1;
				for(int k=0;k<i;k++){
					ll xi=a[j+k],yi=wn*a[i+j+k]%mod;
					a[j+k]=(xi+yi)%mod,a[j+k+i]=(xi-yi+mod)%mod;
					(wn*=w[x])%=mod;
				}
			}
		}
		ll inv=fastpow(n,mod-2);
		if(f<0)for(int i=0;i<n;i++)(a[i]*=inv)%=mod;
	}
}
int sz[maxn],fa[maxn],n;vector<int>graph[maxn];
ll fac[maxn],inv[maxn],b[800010],c[800010],res[maxn];
void dfs(int p,int f){
	sz[p]=1;fa[p]=f;
	for(int i=0;i<graph[p].size();i++){
		int e=graph[p][i];if(e==f)continue;
		dfs(e,p);sz[p]+=sz[e];
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1,u,v;i<n;i++){
		scanf("%d%d",&u,&v);
		graph[u].push_back(v);
		graph[v].push_back(u);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++){
		b[n]++;
		for(int j=0;j<graph[i].size();j++){
			int e=graph[i][j];
			if(e==fa[i]){b[n-sz[i]]--;if(b[n-sz[i]]<0)b[n-sz[i]]+=mod;}
			else {b[sz[e]]--;if(b[sz[e]]<0)b[sz[e]]+=mod;}
		}
	}
	fac[0]=inv[0]=1;
	for(int i=1;i<=n;i++)fac[i]=i*fac[i-1]%mod,(b[i]*=fac[i])%=mod;
	inv[n]=fastpow(fac[n],mod-2);
	for(int i=n-1;i;i--)inv[i]=inv[i+1]*(i+1)%mod;
	for(int i=0;i<=n;i++)c[i]=inv[i];
	for(int i=0;i<=n/2;i++)swap(b[i],b[n-i]);
	ntt::init(n*2-1);ntt::initw(1);
	ntt::ntt(b,1);ntt::ntt(c,1);
	for(int i=0;i<ntt::n;i++)(b[i]*=c[i])%=mod;
	ntt::initw(-1);ntt::ntt(b,-1);
	for(int i=1;i<=n;i++)res[i]=b[n-i]*inv[i]%mod,printf("%lld\n",res[i]);
	return 0;
}