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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=6e5,mod=924844033;
int b[N],n,rev[N],w[N],limn,a[N],siz[N],fst[N],nxt[N],to[N],fac[N],ifac[N],mm;
void ade(int u,int v){to[++mm]=v,nxt[mm]=fst[u],fst[u]=mm;}
void dfs(int u,int fa)
{
    siz[u]=1;++a[n];
    for(int i=fst[u];i;i=nxt[i])
    {
        int v=to[i];if(v==fa)continue;
        dfs(v,u),siz[u]+=siz[v];
    }
    if(u!=1)--a[siz[u]];
    --a[n-siz[u]];
}
int qpower(int a,int b)
{
    int ans=1;for(;b;b>>=1,a=1ll*a*a%mod)if(b&1)ans=1ll*ans*a%mod;
    return ans;
}
void prework(int n)
{
    limn=1;while(limn<=(n<<1))limn<<=1;
    for(int i=1;i<limn;i++)rev[i]=(rev[i>>1]>>1)|((i&1)?(limn>>1):0);
    for(int i=1;i<limn;i<<=1)
    {
        int omg=qpower(5,(mod-1)/(i<<1));
        w[i]=1;
        for(int j=1;j<i;j++)w[i+j]=1ll*w[i+j-1]*omg%mod;
    }
    fac[0]=1;for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
    ifac[n]=qpower(fac[n],mod-2);
    for(int i=n-1;i>=0;i--)ifac[i]=1ll*ifac[i+1]*(i+1)%mod;
}
void DFT(int a[])
{
    for(int i=0;i<limn;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
    for(int i=1;i<limn;i<<=1)
        for(int j=0;j<limn;j+=i<<1)
            for(int k=0;k<i;k++)
            {
                int x=1ll*w[i+k]*a[i+j+k]%mod;
                a[i+j+k]=(a[j+k]-x)%mod,a[j+k]=(a[j+k]+x)%mod;
            }
}
void IDFT(int a[])
{
    reverse(a+1,a+limn);DFT(a);int iv=mod-(mod-1)/limn;
    for(int i=0;i<limn;i++)a[i]=1ll*iv*a[i]%mod;
}
int main()
{
    scanf("%d",&n);
    for(int i=1,u,v;i<n;i++)scanf("%d%d",&u,&v),ade(u,v),ade(v,u);
    dfs(1,0);
    prework(n);
    for(int i=0;i<=n;i++)a[i]=1ll*a[i]*fac[i]%mod,b[i]=ifac[i];
    reverse(a,a+n+1);DFT(a),DFT(b);
    for(int i=0;i<limn;i++)a[i]=1ll*a[i]*b[i]%mod;
    IDFT(a);reverse(a,a+n+1);
    for(int i=1;i<=n;i++)printf("%d\n",(1ll*a[i]*ifac[i]%mod+mod)%mod);
}