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<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<list>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const ll p=924844033;
const ll g=5;
ll fp(ll a,ll b)
{
	ll s=1;
	while(b)
	{
		if(b&1)
			s=s*a%p;
		a=a*a%p;
		b>>=1;
	}
	return s;
}
namespace ntt
{
	ll w1[1000010];
	ll w2[1000010];
	int rev[1000010];
	int n;
	void init()
	{
#ifdef DEBUG
		n=16;
#else
		n=524288;
#endif
		int i;
		for(i=1;i<=n;i<<=1)
		{
			w1[i]=fp(g,(p-1)/i);
			w2[i]=fp(w1[i],p-2);
		}
		rev[0]=0;
		for(i=1;i<n;i++)
			rev[i]=(rev[i>>1]>>1)|(i&1?n>>1:0);
	}
	void ntt(ll *a,int t)
	{
		int i,j,k;
		ll u,v,w,wn;
		for(i=0;i<n;i++)
			if(rev[i]<i)
				swap(a[i],a[rev[i]]);
		for(i=2;i<=n;i<<=1)
		{
			wn=(t==1?w1[i]:w2[i]);
			for(j=0;j<n;j+=i)
			{
				w=1;
				for(k=j;k<j+i/2;k++)
				{
					u=a[k];
					v=a[k+i/2]*w%p;
					a[k]=(u+v)%p;
					a[k+i/2]=(u-v)%p;
					w=w*wn%p;
				}
			}
		}
		if(t==-1)
		{
			ll inv=fp(n,p-2);
			for(i=0;i<n;i++)
				a[i]=a[i]*inv%p;
		}
	}
};
ll b[1000010];
ll c[1000010];
ll a[1000010];
ll inv[1000010];
ll fac[1000010];
ll ifac[1000010];
int s[1000010];
int n;
list<int> l[200010];
ll num[1000010];
void dfs(int x,int fa)
{
	s[x]=1;
	for(auto v:l[x])
		if(v!=fa)
		{
			dfs(v,x);
			s[x]+=s[v];
			num[s[v]]--;
			num[n-s[v]]--;
		}
}
int main()
{
#ifdef DEBUG
	freopen("b.in","r",stdin);
	freopen("b.out","w",stdout);
#endif
	scanf("%d",&n);
	int i,x,y;
	for(i=1;i<=n-1;i++)
	{
		scanf("%d%d",&x,&y);
		l[x].push_back(y);
		l[y].push_back(x);
	}
	inv[0]=inv[1]=fac[0]=fac[1]=ifac[0]=ifac[1]=1;
	for(i=2;i<=n;i++)
	{
		inv[i]=-(p/i)*inv[p%i]%p;
		fac[i]=fac[i-1]*i%p;
		ifac[i]=ifac[i-1]*inv[i]%p;
	}
	dfs(1,0);
	num[n]=n;
	for(i=0;i<=n;i++)
		b[i]=num[n-i]*fac[n-i]%p;
	for(i=0;i<=n;i++)
		c[i]=ifac[i];
	ntt::init();
	ntt::ntt(b,1);
	ntt::ntt(c,1);
	for(i=0;i<ntt::n;i++)
		a[i]=b[i]*c[i]%p;
	ntt::ntt(a,-1);
	for(i=1;i<=n;i++)
	{
		ll ans=a[n-i]*ifac[i]%p;
		ans=(ans+p)%p;
		printf("%lld\n",ans);
	}
	return 0;
}
