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
/* by Natsu Kinmoe */
#include <bits/stdc++.h>
using namespace std;

#define all(x) (x).begin(),(x).end()
#define loop(i,n) for(int i=0;i<(n);i++)
#define cont(i,n) for(int i=1;i<=(n);i++)
#define circ(i,a,b) for(int i=(a);i<=(b);i++)
#define range(i,a,b,c) for(int i=(a);((c)>0?i<=(b):i>=(b));i+=(c))
#define foreach(it,v) for(__typeof((v).begin()) it=(v).begin(),_e_D_=(v).end();it!=_e_D_;it++)
#define y0 y0O0OO00OO0OO0OO0OOO00OO0OO0O0O000OO0
#define y1 y1II11II11III11I1III11II111IIII1II1I1
#define pub push_back
#define pob pop_back
#define mak make_pair

typedef long long ll;
typedef long double lf;
const int Inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3fll;
/* Source code starts here */

const int jt=924844033,rg=5;

int n;
vector<int> nei[1<<19];
int f[1<<19],g[1<<19],sz[1<<19];
int jc[1<<19],fjc[1<<19];
int hoddie[1<<19];

int ksmii(int a,int b){
	if(!b)return 1;
	int x=ksmii(a,b>>1);
	x=1ll*x*x%jt;
	if(b&1)x=1ll*x*a%jt;
	return x;
}

void fft(int a[],int inv){
	loop(i,1<<19)hoddie[i]=((hoddie[i>>1]>>1)|((i&1)<<18));
	loop(i,1<<19)if(i<hoddie[i])swap(a[i],a[hoddie[i]]);
	range(i,1,(1<<19)-1,i){
		int omega=ksmii(rg,(jt-1)+inv*(jt-1)/(i<<1));
		range(j,0,(1<<19)-1,i<<1){
			int omg=1;
			loop(k,i){
				int x=a[j+k],y=1ll*omg*a[j+k+i]%jt;
				a[j+k]=(x+y)%jt;
				a[j+k+i]=(x-y+jt)%jt;
				omg=1ll*omg*omega%jt;
			}
		}
	}
	if(inv==-1){
		int rn=ksmii(1<<19,jt-2);
		loop(i,1<<19)a[i]=1ll*a[i]*rn%jt;
	}
}

void dfs(int now,int lst){
	sz[now]=1;
	foreach(i,nei[now]){
		if(*i==lst)continue;
		dfs(*i,now);sz[now]+=sz[*i];
		f[sz[*i]]++;
	}
	f[n-sz[now]]++;
}

int main(){
	scanf("%d",&n);
	cont(i,n-1){
		int a,b;scanf("%d%d",&a,&b);
		nei[a].pub(b);nei[b].pub(a);
	}
	jc[0]=1;cont(i,n)jc[i]=1ll*jc[i-1]*i%jt;
	fjc[n]=ksmii(jc[n],jt-2);range(i,n,1,-1)fjc[i-1]=1ll*fjc[i]*i%jt;
	dfs(1,0);cont(i,n)f[i]=1ll*f[i]*jc[i]%jt,g[i]=fjc[n-i];
	fft(f,1);fft(g,1);loop(i,1<<19)f[i]=1ll*f[i]*g[i]%jt;fft(f,-1);
	cont(i,n){
		int ans=1ll*n*jc[n]%jt*fjc[i]%jt*fjc[n-i]%jt;
		ans=(ans+jt-1ll*f[n+i]*fjc[i]%jt)%jt;
		printf("%d\n",ans);
	}
	return 0;
}