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 <cstdio>
#include <cctype>
#include <algorithm>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;

typedef long long LL;
const int p=924844033;
int n,inv[200100],s[200100],f[530000],g[530000];
struct edge
{
    int v;
    edge *nxt;
} pool[400100],*tp=pool,*fst[200100];

int getint()
{
    char ch;
    while (!isdigit(ch=getchar()));
    int x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return x;
}
 
int power(int x,int k)
{
    int ret=1;
    for (; k; k>>=1,x=LL(x)*x%p)
        if (k&1)
            ret=LL(ret)*x%p;
    return ret;
}

void ntt(int *a,int n,int flag)
{
    static int rev[530000];
    repu(i,0,n-1)
        if (i<(rev[i]=(rev[i>>1]>>1)|(i&1)*(n>>1)))
            swap(a[i],a[rev[i]]);
    for (int i=1; i<n; i<<=1)
    {
        int wn=power(5,(p-1)/i/2);
        if (flag==-1)
            wn=power(wn,p-2);
        for (int j=0; j<n; j+=i*2)
        {
            int w=1,t;
            repu(k,0,i-1)
            {
                t=LL(w)*a[j+k+i]%p,w=LL(w)*wn%p;
                a[j+k+i]=(a[j+k]+p-t)%p,(a[j+k]+=t)%=p;
            }
        }
    }
    if (flag==1)
        return;
    int inv=power(n,p-2);
    repu(i,0,n-1)
        a[i]=LL(a[i])*inv%p;
}

void dfs(int x,int fa)
{
    s[x]=1;
    for (edge *i=fst[x]; i; i=i->nxt)
        if (i->v!=fa)
            dfs(i->v,x),s[x]+=s[i->v],(f[s[i->v]]+=p-1)%=p;
    ++f[n],(f[n-s[x]]+=p-1)%=p;
}

int main()
{
    n=getint();
    repu(i,1,n-1)
    {
        int u=getint(),v=getint();
        *tp=(edge){v,fst[u]},fst[u]=tp++;
        *tp=(edge){u,fst[v]},fst[v]=tp++;
    }
    dfs(1,0);
    inv[0]=inv[1]=1;
    repu(i,2,n)
        inv[i]=LL(p-p/i)*inv[p%i]%p;
    int t=1;
    repu(i,0,n)
    {
        g[n-i]=inv[i],inv[i+1]=LL(inv[i+1])*inv[i]%p;
        f[i]=LL(f[i])*t%p,t=LL(i+1)*t%p;
    }
    int k=1;
    for (; k<=n*2; k<<=1);
    ntt(f,k,1),ntt(g,k,1);
    repu(i,0,k-1)
        f[i]=LL(f[i])*g[i]%p;
    ntt(f,k,-1);
    repu(i,1,n)
        printf("%d\n",LL(f[i+n])*inv[i]%p);
    return 0;
}