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 <bits/stdc++.h>
using namespace std;
const int maxn=6e5+10,mod=924844033,G=5;
int n,m,k,t,fac[maxn],inv[maxn],ret[maxn];
long long c[maxn],d[maxn],tmp[maxn],wn[maxn];
vector<int>e[maxn];
int qpow(int x,int y)
{
    int ret=1;
    while(y)
    {
        if(y&1)ret=1LL*ret*x%mod;
        x=1LL*x*x%mod;
        y>>=1;
    }
    return ret;
}
void init()
{
    fac[0]=1;
    for(int i=1;i<maxn;i++)fac[i]=1LL*i*fac[i-1]%mod;
    inv[maxn-1]=qpow(fac[maxn-1],mod-2);
    for(int i=maxn-2;i>=0;i--)inv[i]=1LL*inv[i+1]*(i+1)%mod;
}
void getwn(int len)
{
    wn[0]=1;
    wn[1]=qpow(G,(mod-1)/len);
    int i;
    for(i=2;i<=len;i++) wn[i]=wn[i-1]*wn[1]%mod;
}
void NTT(long long *a,int len,int on){
    int i;
    long long ni=qpow(len,mod-2);
    for(i=0;i<len;i++){
        int p=0;
        double ce=log(len)/log(2);
        for(int j=0,tp=i;j<ce;j++,tp/=2) p=(p<<1)+(tp%2);
        tmp[p]=a[i];
    }
    for (int m=2;m<=len;m*=2){
        int half=m/2,bei=len/m;
        for(i=0;i<half;i++){
            long long wi=on>0?wn[i*bei]:wn[len-i*bei];
            for(int j=i;j<len;j+=m){
                long long u=tmp[j],v=tmp[j+half]*wi%mod;
                tmp[j]=(u+v)%mod;
                tmp[j+half]=(u-v+mod)%mod;
            }
        }
    }
    if (on==-1)
        for(i=0;i<len;i++) tmp[i]=tmp[i]*ni%mod;
    for(i=0;i<len;i++) a[i]=tmp[i];
}
int dfs(int x,int y=0)
{
    int sz=1;
    for(int i=0;i<e[x].size();i++)
    {
        int z=e[x][i];
        if(z==y)continue;
        sz+=dfs(z,x);
    }
    c[sz]++;c[n-sz]++;
    return sz;
}
int C(int x,int y)
{
    return 1LL*fac[x]*inv[y]%mod*inv[x-y]%mod;
}
int main()
{
    int i,j;
    init();
    scanf("%d",&n);
    for(i=1;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        e[x].push_back(y),
        e[y].push_back(x);
    }
    dfs(1);
    for(i=0;i<=n;i++)
        c[i]=c[i]*fac[i]%mod,
        d[i]=inv[n-i-1];
    int len=1;
    while(len<n*2)len<<=1;
    getwn(len);
    NTT(c,len,1);
    NTT(d,len,1);
    for(i=0;i<len;i++)c[i]=c[i]*d[i]%mod;
    NTT(c,len,-1);
    for(i=n;i<n<<1;i++)
        printf("%d\n",(1LL*C(n,i-n+1)*(n+1)%mod-c[i]*inv[i-n+1]%mod+mod)%mod);
    return 0;
}