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<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#define ll long long
using namespace std;
const int MAXN=200233;
const int D=924844033,G=5,iG=554906420;
int n,ans[MAXN];
vector<int> ver[MAXN];

int fac[MAXN],inv[MAXN];
int rev[MAXN<<2];
int A[MAXN<<2],B[MAXN<<2];

int qpow(int x,int y) {
	int ret=1;
	while(y) {
		if(y&1) ret=1ll*ret*x%D;
		x=1ll*x*x%D;
		y>>=1;
	}
	return ret;
}

void NTT(int f[],int op,int lim) {
	for(int i=1;i<lim;i++)
		if(rev[i]>i) swap(f[rev[i]],f[i]);
	for(int i=1;i<lim;i<<=1) {
		int sta=qpow(~op?G:iG,(D-1)/(i<<1));
		for(int j=0;j<lim;j+=(i<<1)) {
			int w=1;
			for(int k=0;k<i;k++,w=1ll*w*sta%D) {
				int t1=f[j+k],t2=1ll*w*f[i+j+k]%D;
				f[j+k]=(t1+t2)%D;
				f[i+j+k]=(t1-t2+D)%D;
			} 
		}
	}
	if(~op) return ;
	int ilim=qpow(lim,D-2);
	for(int i=0;i<lim;i++)
		f[i]=1ll*f[i]*ilim%D;
}

int siz[MAXN];
void dfs(int u,int fa) {
	siz[u]=1;
	for(int v:ver[u])
		if(v!=fa) dfs(v,u),siz[u]+=siz[v];
}

void first() {
	fac[0]=1;
	for(int i=1;i<=n;i++) fac[i]=1ll*i*fac[i-1]%D;
	inv[n]=qpow(fac[n],D-2);
	for(int i=n-1;~i;i--) inv[i]=1ll*inv[i+1]*(i+1)%D;
}
int C(int x,int y) {
	return 1ll*fac[x]*inv[y]%D*inv[x-y]%D;
}

int main() {
	scanf("%d",&n);
	first();
	for(int i=1;i<n;i++) {
		int x,y; scanf("%d%d",&x,&y);
		ver[x].push_back(y);
		ver[y].push_back(x);
	}
	dfs(1,0);
	siz[0]=1;
	for(int i=2;i<=n;i++)
		++A[siz[i]],++A[n-siz[i]];
	for(int i=0;i<=n;i++)
		A[i]=1ll*A[i]*fac[i]%D;
	for(int i=0;i<=n;i++)
		B[i]=inv[i];
	reverse(B,B+n+1);

	int lim; for(lim=1;lim<=(n<<1);lim<<=1);
	for(int i=1;i<lim;i++)
		rev[i]=(rev[i>>1]>>1)|(i&1?lim>>1:0);
	NTT(A,1,lim),NTT(B,1,lim);
	for(int i=0;i<lim;i++)
		A[i]=1ll*A[i]*B[i]%D;
	NTT(A,-1,lim);

	for(int i=1;i<=n;i++)
		ans[i]=((1ll*C(n,i)*n-1ll*inv[i]*A[n+i])%D+D)%D;

	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
	return 0;
}