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<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int N=200005;
typedef long long ll;
const ll mod=924844033;

int n,hd[N],to[N*2],pr[N*2],tot,sz[N];
void addedge(int u,int v) {to[++tot]=v;pr[tot]=hd[u];hd[u]=tot;}
void dfs(int u,int fa) 
{
	sz[u]=1;
	for (int i=hd[u];i;i=pr[i]) if (to[i]!=fa) dfs(to[i],u),sz[u]+=sz[to[i]];
}



int jc[N],ny[N],ans[N];
int C(int n,int m) 
{
	if (n<m) return 0;
	return (ll)jc[n]*ny[m]%mod*ny[n-m]%mod;
}
int power_mod(int a,int b) 
{
	int ans=1;
	while (b) 
	{
		if (b&1) ans=(ll)ans*a%mod;
		a=(ll)a*a%mod;
		b>>=1;
	}
	return ans;
}
void MOD(int &x) {if (x>=mod) x-=mod;}
int om[N*4],on[N*4],rev[N*4],len;
void init(int n)
{
	int l=0;len=1;while (len<n) len<<=1,l++;
	int g=power_mod(5,(mod-1)/len),inv=power_mod(g,mod-2);
	for (int i=0;i<len;i++) 
	{
		om[i]= i==0 ? 1:(ll)om[i-1]*g%mod;
		on[i]= i==0 ? 1:(ll)on[i-1]*inv%mod;
		rev[i]=(rev[i>>1]>>1)|((i&1)<<l-1);
	}
}
void ntt(int *a,int *om,int n) 
{
	for (int i=0;i<n;i++) if (rev[i]>i) swap(a[i],a[rev[i]]);
	for (int len=2;len<=n;len<<=1) 
		for (int m=len/2,i=0;i<n;i+=len)
			for (int t,j=0;j<m;j++) 
			{
				t=(ll)a[i+j+m]*om[n/len*j]%mod;
				MOD(a[i+j+m]=a[i+j]-t+mod);
				MOD(a[i+j]+=t);
			}
}
void dft(int *a) {ntt(a,om,len);}
void idft(int *a) 
{
	ntt(a,on,len);
	int inv=power_mod(len,mod-2);
	for (int i=0;i<len;i++) a[i]=(ll)a[i]*inv%mod;
}
int c[N*4],d[N*4];


int main()
{
	scanf("%d",&n);
	for (int u,v,i=1;i<n;i++) 
	{
		scanf("%d%d",&u,&v);
		addedge(u,v);
		addedge(v,u);
	}
	dfs(1,0);
	
	//for (int i=1;i<=n;i++) printf("sz[%d]=%d\n",i,sz[i]);
	
	jc[0]=jc[1]=ny[0]=ny[1]=1;
	for (int i=2;i<=n;i++) jc[i]=(ll)jc[i-1]*i%mod,ny[i]=(mod-mod/i)*ny[mod%i]%mod;
	for (int i=2;i<=n;i++) ny[i]=(ll)ny[i-1]*ny[i]%mod;
	
	
	init(n*2+1);	
	for (int i=2;i<=n;i++) c[sz[i]]++;
	for (int i=2;i<=n;i++) c[n-sz[i]]++; 
	for (int i=n;i>=0;i--) d[i]=ny[n-i],c[i]=(ll)c[i]*jc[i]%mod;
	//for (int i=0;i<len;i++) printf("i=%d d=%d c=%d\n",i,d[i],c[i]);
	dft(c),dft(d);
	for (int i=0;i<len;i++) d[i]=(ll)d[i]*c[i]%mod;
	idft(d);
	for (int i=n+1;i<=n*2;i++) ans[i-n]=d[i];//printf("ans[%d]=%d\n",i-n,ans[i-n]);
	
	
	
	for (int i=1;i<=n;i++) 
	printf("%lld\n",(( (ll)n*C(n,i)-(ll)ans[i]*ny[i] )%mod+mod)%mod   );
//	system("pause");
	return 0;
}