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 <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
#define pb push_back
using namespace std;
const int P=924844033,G=5,iG=554906420,maxn=8e5+10;
int n,r[maxn],fac[maxn],ifac[maxn],sz[maxn],N,l,u,v;vector<int> H[maxn];
typedef vector<int> Poly;
inline int Add(int x,int y){x+=y;return x>=P?x-P:x;}
inline int Dec(int x,int y){x-=y;return x<0?x+P:x;}
inline int Mul(int x,int y){return 1ll*x*y%P;}
int ksm(int a,int b){
	int ans=1;
	for(;b;b>>=1,a=Mul(a,a)) if(b&1) ans=Mul(ans,a);
	return ans;
}
void DFT(Poly &a,int k){
	rep(i,0,N-1){
		r[i]=(i&1)?(r[i^1]|(1<<l-1)):(r[i>>1]>>1);
		if(i<r[i]) swap(a[i],a[r[i]]);
	}
	for(int mid=1;mid<N;mid<<=1){
		int wn=ksm(k==1?G:iG,(P-1)/(mid<<1));
		for(int R=mid<<1,j=0;j<N;j+=R){
			int w=1,*l=&a[j],*r=&a[j+mid];
			for(int k=0;k<mid;++k,++l,++r,w=Mul(w,wn)){
				int x=*l,y=Mul(*r,w);
				*l=Add(x,y);*r=Dec(x,y);
			}
		}
	}
	if(k==-1){
		int inv=ksm(N,P-2);
		rep(i,0,N-1) a[i]=Mul(a[i],inv);
	}
}
Poly operator *(Poly a,Poly b){
	int n=a.size()+b.size()-1;
	for(N=1,l=0;N<=n;N<<=1,l++);a.resize(N);b.resize(N);
	DFT(a,1);DFT(b,1);
	rep(i,0,N-1) a[i]=Mul(a[i],b[i]);
	DFT(a,-1);a.resize(n);
	return a;
}
void pre(int n){
	fac[0]=1;
	rep(i,1,n) fac[i]=Mul(fac[i-1],i);
	ifac[n]=ksm(fac[n],P-2);
	per(i,n-1,0) ifac[i]=Mul(ifac[i+1],i+1);
}Poly a,b;
void dfs(int u,int fa){
	sz[u]=1;
	for(int v:H[u]) if(v^fa) dfs(v,u),sz[u]+=sz[v],++a[sz[v]];
	++a[n-sz[u]];
}
inline int C(int n,int k){return Mul(fac[n],Mul(ifac[k],ifac[n-k]));}
int main(){
	scanf("%d",&n);pre(n);a.resize(n+1);b.resize(n+1);
	rep(i,1,n-1) scanf("%d%d",&u,&v),H[u].pb(v),H[v].pb(u);
	dfs(1,0);
	rep(i,1,n) a[i]=Mul(a[i],fac[i]);reverse(a.begin(),a.end());
	rep(i,0,n) b[i]=ifac[i];
	a=a*b;a.resize(n+1);reverse(a.begin(),a.end());
	rep(i,1,n) printf("%d\n",Dec(Mul(C(n,i),n),Mul(ifac[i],a[i])));
	return 0;
}