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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int Mod=924844033,g=5,gi=554906420;
const int N=200005;
inline int qpow(int a,int b){
	int ans=1;
	while (b){
		if (b&1) ans=1ll*ans*a%Mod;
		a=1ll*a*a%Mod,b>>=1;
	}
	return ans;
}
int rev[N<<2],a[N<<2],b[N<<2],lim;
inline void NTT_init(int n){
	int l=0;for (lim=1;lim<=(n<<1);lim<<=1) ++l;
	for (int i=1;i<=lim;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
}
inline void NTT(int*s,int t){
	for (int i=0;i<lim;i++) if (i<rev[i]) swap(s[i],s[rev[i]]);
	for (int mid=1;mid<lim;mid<<=1){
		int wn=qpow(t==1?g:gi,(Mod-1)/(mid<<1));
		for (int i=0;i<lim;i+=(mid<<1)){
			int w=1;
			for (int j=0;j<mid;j++,w=1ll*w*wn%Mod){
				int x=s[i+j],y=1ll*s[i+j+mid]*w%Mod;
				s[i+j]=(x+y)%Mod,s[i+j+mid]=(x-y+Mod)%Mod;
			}
		}
	}
	if (t==-1) {
		int inv=qpow(lim,Mod-2);
		for (int i=0;i<=lim;i++) s[i]=1ll*s[i]*inv%Mod;
	}
}
int size[N],cnt[N],fac[N],fac_inv[N];
int n;
inline void binom_init(){
	fac[0]=1;
	for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%Mod;
	fac_inv[n]=qpow(fac[n],Mod-2);
	for (int i=n-1;i>=0;i--) fac_inv[i]=1ll*fac_inv[i+1]*(i+1)%Mod; 
}
inline int C(int x,int y){
	return 1ll*fac[x]*fac_inv[y]%Mod*fac_inv[x-y]%Mod;
}
int Head[N],Next[N<<1],Adj[N<<1],tot=0;
inline void addedge(int u,int v){
	Next[++tot]=Head[u];
	Head[u]=tot;
	Adj[tot]=v;
	Next[++tot]=Head[v];
	Head[v]=tot;
	Adj[tot]=u;
}
inline void dfs(int x,int f){
	size[x]=1;
	for (int e=Head[x];e;e=Next[e])
		if (Adj[e]!=f){
			dfs(Adj[e],x);
			size[x]+=size[Adj[e]];
		}
	if (f) cnt[size[x]]++,cnt[n-size[x]]++;
}
int f[N];
int main (){
	scanf ("%d",&n);
	for (int i=1,u,v;i<n;i++){
		scanf ("%d%d",&u,&v);
		addedge(u,v);
	}
	dfs(1,0);
	binom_init();
	NTT_init(n);
	for (int i=1;i<=n;i++) a[i]=1ll*cnt[i]*fac[i]%Mod;
    for (int i=0;i<=n;i++) b[n-i]=fac_inv[i];
    NTT(a,1),NTT(b,1);
    for (int i=0;i<=lim;i++) a[i]=1ll*a[i]*b[i]%Mod;
    NTT(a,-1);
    for (int i=1;i<=n;i++) f[i]=1ll*a[n+i]*fac_inv[i]%Mod;
    for (int i=1;i<=n;i++) printf ("%d\n",(1ll*n*C(n,i)%Mod-f[i]+Mod)%Mod);
	return 0;
}
