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<algorithm>
using namespace std;
typedef long long ll;
const int N=200200,M=1<<19,P=924844033;
int i,j,k,n,m,ch,ff,En,x,y;
int a[N],sz[N],h[N],Jc[N],Jc_[N];
int A[M],B[M],rev[M],W[2][M];
struct edge { int s,n;} E[N<<1];
void R(int &x) {
	ff=x=0;ch=getchar();
	while (ch<'0' || '9'<ch) { if (ch=='-') ff=1;ch=getchar();}
	while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar();
	if (ff) x=-x;
}
int ksm(int x,int y) {
	int z=1;
	for (;y;y>>=1,x=(ll) x*x%P) if (y&1) z=(ll) z*x%P;
	return z;
}
void pre(int n) {
	int i;
	Jc[0]=Jc_[0]=1; 
	for (i=1;i<=n;i++) Jc[i]=(ll) Jc[i-1]*i%P;
	Jc_[n]=ksm(Jc[n],P-2);
	for (i=n-1;i;i--) Jc_[i]=(ll) Jc_[i+1]*(i+1)%P;
}
void Pre(int n,int m) {
	int i,j,k;
	for (i=0;i<n;i++) {
		k=i;
		for (j=0;j<m;j++) rev[i]=(rev[i]<<1)|(k&1),k>>=1;
	}
	int w=ksm(5,(P-1)/n);
	W[0][0]=W[1][0]=1;
	for (i=1;i<n;i++) W[0][i]=(ll) w*W[0][i-1]%P;
	for (i=1;i<n;i++) W[1][i]=W[0][n-i];
}
void E_add(int x,int y) {
	E[++En].s=y;E[En].n=h[x];h[x]=En;
	E[++En].s=x;E[En].n=h[y];h[y]=En;
}
void dfs(int x,int F) {
	sz[x]=1;
	for (int k=h[x];k;k=E[k].n) if (E[k].s!=F) {
		dfs(E[k].s,x);
		sz[x]+=sz[E[k].s];
		a[sz[E[k].s]]++;
	}
	if (sz[x]!=n) a[n-sz[x]]++;
}
void NTT(int *A,int n,int m,int fg) {
	int i,j,k,l,t,u;
	for (i=0;i<n;i++) if (i<rev[i]) swap(A[i],A[rev[i]]);
	for (i=1;i<n;i<<=1)
		for (j=0;j<n;j+=i<<1)
			for (k=0,l=0;k<i;k++,l+=n/(i<<1)) {
				t=(ll) W[fg][l]*A[i+j+k]%P;
				u=A[j+k];
				A[j+k]=(u+t)%P;
				A[i+j+k]=(u+P-t)%P;
			}
	int n_=ksm(n,P-2);
	if (fg) for (i=0;i<n;i++) A[i]=(ll) n_*A[i]%P;
}
int main() {
	R(n);
	pre(n);
	for (i=1;i<n;i++) {
		R(x);R(y);
		E_add(x,y);
	}
	dfs(1,0);
	n++;
	for (i=1;i<n;i++) A[i]=(ll) Jc[i]*a[i]%P;
	for (i=0;i+i+1<n;i++) swap(A[i],A[n-1-i]);
	for (i=0;i<n;i++) B[i]=Jc_[i];
	m=0;
	while ((1<<m)<n) m++;
	m++;
	Pre(1<<m,m);
	NTT(A,1<<m,m,0);
	NTT(B,1<<m,m,0);
	for (i=0;i<(1<<m);i++) A[i]=(ll) A[i]*B[i]%P; 
	NTT(A,1<<m,m,1);
	for (i=0;i+i+1<n;i++) swap(A[i],A[n-1-i]);
	for (i=1;i<n;i++) {
		int t=(ll) Jc_[i]*A[i]%P;
		t=((ll) Jc[n-1]*Jc_[i]%P*Jc_[n-1-i]%P*(n-1)+P-t)%P;
		printf("%d\n",t);
	}
}