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<iostream>
#include<cstdio>
using namespace std;
#define ll long long
const ll p=924844033,g=5,gv=554906420,N=1e6+5;
ll qpow(ll a,int b)
{ll ret=1;while(b){if(b&1)ret=ret*a%p;a=a*a%p;b>>=1;}return ret;}
ll inv(ll a){return qpow(a,p-2);}
int rev[N];
void getrev(int len)
{for(int i=0;i<len;i++){rev[i]=rev[i>>1]>>1;if(i&1)rev[i]|=len>>1;}}
void ntt(ll x[],int len,int form)
{
	int i,j,k,step;ll s1,s2,w,wn,lv;
	for(i=0;i<len;i++)if(i<rev[i])swap(x[i],x[rev[i]]);
	for(i=2;i<=len;i<<=1)
	{
		step=i>>1;wn=qpow(form==1?g:gv,(p-1)/i);
		for(j=0;j<len;j+=i)
		{
			w=1;
			for(k=j;k<j+step;k++)
			{
				s1=x[k],s2=x[k+step]*w%p;
				x[k]=(s1+s2)%p,x[k+step]=(s1-s2+p)%p,w=w*wn%p;
			}
		}
	}
	if(form==-1){lv=inv(len);for(i=0;i<len;i++)x[i]=x[i]*lv%p;}
}
struct edge{int to,nxt;}e[N<<1];
int head[N],tot=1,sz[N],fa[N];
void adde(int x,int y){e[tot]=(edge){y,head[x]};head[x]=tot++;}
void dfs(int u,int f)
{
    sz[u]=1;fa[u]=f;
    for(int i=head[u];i;i=e[i].nxt)
        if(e[i].to^f)dfs(e[i].to,u),sz[u]+=sz[e[i].to];
}
ll a[N],b[N],fac[N],finv[N];
ll comb(int n,int m){return m>n?0:fac[n]*finv[n-m]%p*finv[m]%p;}
int tt=0;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<n;i++)
        {int x,y;scanf("%d%d",&x,&y);adde(x,y),adde(y,x);}
    fac[0]=finv[0]=1;
    for(int i=1;i<=2*n+5;i++)fac[i]=fac[i-1]*i%p,finv[i]=inv(fac[i]);
    dfs(1,1);
    for(int i=1;i<=n;i++)
    {
        if(i!=1)a[n-sz[i]]++;
        for(int j=head[i];j;j=e[j].nxt)a[sz[e[j].to]]+=(fa[i]!=e[j].to);
    }
    for(int i=0;i<=n;i++)a[i]=a[i]*fac[i]%p,b[i]=finv[n-i];
    //for(int i=0;i<=n;i++)cout<<a[i]<<" "<<b[i]<<endl;
    int len=1;while(len<=(2*n))len<<=1;
    getrev(len);ntt(a,len,1),ntt(b,len,1);
    for(int i=0;i<len;i++)a[i]=a[i]*b[i]%p;
    ntt(a,len,-1);
    for(int k=1;k<=n;k++)
        printf("%lld\n",((n*comb(n,k)-finv[k]*a[n+k])%p+p)%p);
}
