Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<vector>
#define maxn 2000010
#define ll long long
#define mod 924844033
using namespace std;
int n,x,y,cnt[maxn],siz[maxn],r[maxn];
ll fac[maxn],inv[maxn],A[maxn],B[maxn];
vector<int>vec[maxn];
ll ksm(ll x,ll i){
	ll res=1;
	for(;i;i>>=1,x=x*x%mod)
		if(i&1)res=res*x%mod;
	return res;
}
void dfs(int u,int fa){
	siz[u]=1;
	for(auto v:vec[u]){
		if(v==fa)continue;
		dfs(v,u);
		siz[u]+=siz[v];
	}
	cnt[siz[u]]--;
	if(u!=1)cnt[n-siz[u]]--;
}
inline void NTT(ll *A,int lim,int opt){
	for(register int i=0;i<lim;i++)if(i<r[i])swap(A[i],A[r[i]]);
	for(register int mid=1;mid<lim;mid<<=1){
		ll W=ksm(5,(mod-1)/(mid<<1));if(opt==-1)W=ksm(W,mod-2);
		for(register int step=mid<<1,j=0;j<lim;j+=step){
			ll w=1;
			for(register int k=0;k<mid;k++,w=w*W%mod){
				ll x=A[j+k],y=w*A[j+mid+k]%mod;
				A[j+k]=(x+y)%mod;
				A[j+mid+k]=(x-y+mod)%mod;
			}
		}
	}
	if(opt==-1){
		ll inv=ksm(lim,mod-2);
		for(register int i=0;i<lim;i++)A[i]=A[i]*inv%mod;
	}
}
int main(){
	scanf("%d",&n);
	fac[0]=inv[0]=1;
	for(register int i=1;i<=n;i++)fac[i]=fac[i-1]*i%mod;
	inv[n]=ksm(fac[n],mod-2);
	for(register int i=n-1;i;i--)inv[i]=inv[i+1]*(i+1)%mod;
	for(register int i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		vec[x].push_back(y);
		vec[y].push_back(x);
	}
	dfs(1,0);
	cnt[n]=n;
	int lim=1,l=0;
	while(lim<n*2)lim<<=1,l++;
	for(register int i=0;i<lim;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	for(register int i=0;i<=n;i++)A[n-i]=fac[i]*cnt[i]%mod;
	for(register int i=0;i<=n;i++)B[i]=inv[i];
	NTT(A,lim,1);NTT(B,lim,1);
	for(register int i=0;i<lim;i++)A[i]=A[i]*B[i]%mod;
	NTT(A,lim,-1);
	for(register int i=1;i<=n;i++)printf("%lld\n",(A[n-i]*inv[i]%mod+mod)%mod);
	return 0;
}