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 <algorithm>
#include <cstdio>

const int N=200005,M=1<<20,mo=924844033;
int a[M],b[M],rev[M],g[20][2],fst[N],fac[N],inv[N],n,tot=0,invn;
struct Edge{
	int to,nxt;
}e[N*2];

void addedge(int x,int y){
	e[++tot].to=y;e[tot].nxt=fst[x];fst[x]=tot;
}

int dfs(int x,int father){
	int sz=1;
	for (int i=fst[x];i;i=e[i].nxt)
		if (e[i].to!=father) sz+=dfs(e[i].to,x);
	if (x!=1) a[sz]++,a[n-sz]++;
	return sz;
}

void FFT(int *a,int n,int f){
	for (int i=0;i<n;i++)
		if (i<rev[i]) std::swap(a[i],a[rev[i]]);
	for (int i=1,t=0;i<n;i<<=1,t++){
		int wn=g[t][f];
		for (int j=0;j<n;j+=i<<1){
			int w=1;
			for (int k=0;k<i;k++,w=1ll*w*wn%mo){
				int x=a[j+k],y=1ll*w*a[j+k+i]%mo;
				a[j+k]=(x+y)%mo;
				a[j+k+i]=(x-y)%mo;
			}
		}
	}
	if (f)
		for (int i=0;i<n;i++) a[i]=1ll*a[i]*invn%mo;
}

int ksm(int x,int y){
	int i=1;
	for (;y;y>>=1,x=1ll*x*x%mo)
		if (y&1) i=1ll*i*x%mo;
	return i;
}

void mul(int m){
	int L=0,n=1;
	for (;n<m;n<<=1) L++;
	invn=ksm(n,mo-2);
	for (int i=1;i<n;i++)
		rev[i]=rev[i>>1]>>1^((i&1)<<(L-1));
	int val0=ksm(5,(mo-1)/n),val1=ksm(val0,mo-2);
	for (int i=L-1;i>=0;i--){
		g[i][0]=val0;val0=1ll*val0*val0%mo;
		g[i][1]=val1;val1=1ll*val1*val1%mo;
	}
	FFT(a,n,0);FFT(b,n,0);
	for (int i=0;i<n;i++) a[i]=1ll*a[i]*b[i]%mo;
	FFT(a,n,1);
}

int main(){
	scanf("%d\n",&n);
	for (int i=1;i<n;i++){
		int x,y;
		scanf("%d%d\n",&x,&y);
		addedge(x,y);
		addedge(y,x);
	}
	dfs(1,0);
	fac[0]=1;
	for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mo;
	inv[n]=ksm(fac[n],mo-2);
	for (int i=n;i;i--) inv[i-1]=1ll*inv[i]*i%mo;
	for (int i=0;i<=n;i++){
		a[i]=1ll*a[i]*fac[n-i]%mo;
		b[i]=inv[i];
	}
	mul(n*2);
	for (int i=1;i<=n;i++){
		int ans=(1ll*fac[n]*inv[n-i]%mo*n-a[n-i])%mo*inv[i]%mo;
		printf("%d\n",(ans+mo)%mo);
	}
}