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>
using namespace std;
const int maxn=8e5+9,p=924844033;
int r[maxn];
inline int poww(int a,int b){
    int ans=1;
    while(b){
        if(b&1)ans=1ll*ans*a%p;
        b>>=1;
        a=1ll*a*a%p;
    }
    return ans%p;
}
inline int inv(int x){
    return poww(x,p-2)%p;
}
inline int getrev(int n,int m){
    int lim=1,l=0;
    while(lim<=n+m)++l,lim<<=1;
    for(int i=0;i<=lim;++i)r[i]=r[i>>1]>>1|(i&1)<<l-1;
    return lim;
}
inline void NTT(int limit,int *A,int type){
    int pr=5;
    for(int i=0;i<limit;++i)if(r[i]>i)swap(A[i],A[r[i]]);
    for(int mid=1;mid<limit;mid<<=1){
        int wn=poww(type==-1?inv(pr):pr,(p-1)/(mid<<1));
        for(int j=0,R=mid<<1;j<limit;j+=R){
            int w=1;
            for(int k=0;k<mid;++k,w=1ll*w*wn%p){
                int x=A[j+k],y=1ll*w*A[j+k+mid]%p;
                A[j+k]=(x+y)%p;
                A[j+k+mid]=(x-y+p)%p;
            }
        }
    }
    if(type==-1){
        int INV=inv(limit);
        for(int i=0;i<limit;++i)A[i]=1ll*A[i]*INV%p;    
    }
}
vector<int>v[maxn];
int cnt[maxn],n,fac[maxn],ifac[maxn],siz[maxn];
void dfs(int x,int fa){
	siz[x]=1;
	for(auto y:v[x]){
		if(y==fa)continue;
		dfs(y,x);
		siz[x]+=siz[y];
		++cnt[siz[y]];
	}
	++cnt[n-siz[x]];
}
int a[maxn],b[maxn];
int main(){
	cin>>n;
	fac[0]=1;for(int i=1;i<=n;++i)fac[i]=1ll*fac[i-1]*i%p;
	ifac[n]=inv(fac[n]);for(int i=n-1;i>=0;--i)ifac[i]=1ll*ifac[i+1]*(i+1)%p;
	for(int i=1,x,y;i<n;++i){
		cin>>x>>y;v[x].push_back(y);v[y].push_back(x);
	}
	dfs(1,0);cnt[0]=0;
	for(int i=1;i<=n;++i)a[i]=1ll*fac[i]*cnt[i]%p;
	for(int i=0;i<=n;++i)b[i]=ifac[i];
	reverse(a,a+n+1);
	int lim=getrev(n,n);
	NTT(lim,a,1);NTT(lim,b,1);
	for(int i=0;i<=lim;++i)a[i]=1ll*a[i]*b[i]%p;
	NTT(lim,a,-1);
	reverse(a,a+n+1);
	for(int i=1;i<=n;++i)a[i]=1ll*a[i]*ifac[i]%p,a[i]=(1ll*n*fac[n]%p*ifac[i]%p*ifac[n-i]%p+p-a[i])%p;
	for(int i=1;i<=n;++i)cout<<a[i]<<endl;
}