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<bits/stdc++.h>
#define ll long long
#define inf(x) (ll)(1e##x)
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
const int mod = 924844033,N = 1001000;
namespace FFT{
	const int g = 5;
	int len,maxn,r[N];
	ll g1[N],g2[N];
	ll fp(ll a,ll k){
		ll ans=1;
		for(;k;k>>=1,a=(a*a)%mod)
			if(k&1) ans=(a*ans)%mod;
		return ans;
	}
	void build(int n){
		maxn=1;len=0;while(maxn<=n) maxn<<=1,len++;
		for(int i=0;i<maxn;i++) r[i]=r[i>>1]>>1|(i&1)<<len-1;
	}
	void NTT(ll *a,int tag){
		for(int i=0;i<maxn;i++)
			if(i<r[i]) swap(a[i],a[r[i]]);
		for(int mid=1;mid<maxn;mid<<=1){
			ll wn=fp(g,(mod-1)/mid/2);
			if(tag<0) wn=fp(wn,mod-2);
			for(int l=0;l<maxn;l+=mid<<1)
				for(int i=l,w=1;i<l+mid;i++,w=wn*w%mod){
					ll x=a[i],y=a[i+mid]*w;
					a[i]=(x+y)%mod; a[i+mid]=(x-y)%mod;
				}
		}
	}
	void mul(ll *a,ll *b){for(int i=0;i<maxn;i++) a[i]=a[i]*b[i]%mod;}
	void clear(ll *a,int bg=0){for(int i=bg;i<maxn;i++) a[i]=0;}
	void turn(ll *a,ll *b,int ed=maxn){memcpy(b,a,8*ed+8);}
	void iturn(ll *a,ll *b){ll inv=fp(maxn,mod-2);for(int i=0;i<maxn;i++) b[i]=a[i]*inv%mod;}
	void mult(ll *a,ll *b,ll *c,int n,int m){
		build(n+m);clear(g1);clear(g2);
		turn(a,g1,n);turn(b,g2,m);
		NTT(g1,1);NTT(g2,1);mul(g1,g2);NTT(g1,-1);iturn(g1,c);
	}
}
ll f[N],g[N],ifac[N],fac[N];
int s[N],n,h[N];
vector<int> e[N];
void dfs(int x,int dad){
	s[x]=1;
	for(int v:e[x]) if(v!=dad){
		dfs(v,x);s[x]+=s[v];
		h[s[v]]++;
	}
	if(dad) h[n-s[x]]++;
}
int main(){
	read(n);
	for(int i=1;i<n;i++){
		int x,y;read(x);read(y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1,0);
	fac[0]=1;
	for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
	ifac[n]=FFT::fp(fac[n],mod-2);
	for(int i=n;i;i--) ifac[i-1]=ifac[i]*i%mod;
	for(int i=0;i<=n;i++) g[i]=FFT::fp(fac[n-i],mod-2);
	for(int i=0;i<=n;i++) f[i]=fac[i]*h[i]%mod;
	FFT::mult(f,g,f,n,n);
	for(int i=1;i<=n;i++){
		ll ans=fac[n]*ifac[i]%mod*ifac[n-i]%mod*n%mod;
		ans=(ans-f[n+i]*ifac[i])%mod;
		cout<<(ans+mod)%mod<<'\n';
	}
	return 0;
}