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<algorithm>
#include<iostream>
#include<cstdio>
#include<vector>
#define ll long long
#define poly vector<ll>
using namespace std;
const int N=5250007;
const int mod=924844033;
int n;
int rev[N],g[N],sz[N];
vector<int> ve[N];
ll fac[N],finv[N];
poly f,h;
inline ll qpow(ll x,int k,ll r=1){
	for(;k;k>>=1,x=x*x%mod) if(k&1) r=r*x%mod;
	return r;
}
inline void ntt(poly &a,int len,int opt){
	a.resize(len);
	for(int i=0;i<len;++i) if(i<rev[i]) swap(a[i],a[rev[i]]);
	for(int i=1;i<len;i<<=1) for(int j=0;j<len;j+=i<<1) for(int k=0;k<i;++k){
		int x=a[j+k],y=a[j+k+i]*g[i+k]%mod;
		a[j+k]=x+y>=mod?x+y-mod:x+y; a[j+k+i]=x-y<0?x-y+mod:x-y;
	}
	if(~opt) return ;
	reverse(a.begin()+1,a.end());
	const int inv=qpow(len,mod-2);
	for(int i=0;i<len;++i) a[i]=a[i]*inv%mod;
}
void dfs(int x,int fa){
	sz[x]=1;
	for(auto to:ve[x]) if(to!=fa) dfs(to,x),sz[x]+=sz[to];
	++f[n]; if(fa) --f[sz[x]],--f[n-sz[x]];
}
int main(){
	scanf("%d",&n); int len=1,bit=0;
	for(;len<=n+n;len<<=1,++bit);
	for(int i=0;i<len;++i) rev[i]=rev[i>>1]>>1|(i&1)<<bit-1;
	for(int i=fac[0]=1;i<=n;++i) fac[i]=fac[i-1]*i%mod;
	finv[n]=qpow(fac[n],mod-2);
	for(int i=n;i;--i) finv[i-1]=finv[i]*i%mod;
	for(int i=1,a,b;i<n;++i) scanf("%d%d",&a,&b),ve[a].push_back(b),ve[b].push_back(a);
	for(int i=len/2,wn=qpow(5,(mod-1)/len),w=1;i<len;++i,w=1ll*w*wn%mod) g[i]=w;
	for(int i=len/2-1;i;--i) g[i]=g[i<<1];
	f.resize(n+1); h.resize(n+1); dfs(1,0);
	for(int i=1;i<=n;++i) f[i]=(f[i]+mod)*fac[i]%mod;
	for(int i=0;i<=n;++i) h[i]=finv[n-i];
	ntt(f,len,1); ntt(h,len,1);
	for(int i=0;i<len;++i) f[i]=f[i]*h[i]%mod;
	ntt(f,len,-1);
	for(int i=1;i<=n;++i) printf("%lld\n",f[i+n]*finv[i]%mod);
	return 0;
}