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 N=8e5+50,mod=924844033,G=5,Gi=554906420;
int n,ver[N*2],nxt[N*2],head[N],tot,sz[N],b[N],a[N],J[N],I[N],lim=1,lo=-1,r[N];
void add(int x,int y){ver[++tot]=y;nxt[tot]=head[x];head[x]=tot;}
void dfs(int x,int ff){
    sz[x]=1;
    for(int i=head[x],y;i;i=nxt[i])
        if((y=ver[i])!=ff)dfs(y,x),sz[x]+=sz[y],b[sz[y]]++;
    b[n-sz[x]]++;
}
int power(int x,int y){
    int z=1;
    for(;y;y>>=1,x=1ll*x*x%mod)if(y&1)z=1ll*z*x%mod;
    return z;
}
void NTT(int *a,int op){
    for(int i=0;i<lim;i++)if(i<r[i])swap(a[i],a[r[i]]);
    for(int i=1;i<lim;i<<=1)
        for(int t=i<<1,j=0,Wn=power(op>0?G:Gi,(mod-1)/t);j<lim;j+=t)
            for(int k=0,w=1;k<i;k++,w=1ll*w*Wn%mod){
                int x=a[j+k],y=1ll*a[i+j+k]*w%mod;
                a[j+k]=(x+y)%mod;a[i+j+k]=(x-y)%mod;
            }
    if(op<0)for(int i=0,inv=power(lim,mod-2);i<lim;i++)a[i]=1ll*a[i]*inv%mod;
}
int C(int n,int m){return 1ll*J[n]*I[m]%mod*I[n-m]%mod;}
int main(){
    scanf("%d",&n);J[0]=I[0]=I[1]=1;
    for(int i=2;i<=n;i++)I[i]=mod-1ll*mod/i*I[mod%i]%mod;
    for(int i=1;i<=n;i++)J[i]=1ll*J[i-1]*i%mod,I[i]=1ll*I[i-1]*I[i]%mod;
    for(int i=1,x,y;i<n;i++){
        scanf("%d%d",&x,&y);
        add(x,y);add(y,x);
    }
    dfs(1,0);
    for(int i=1;i<=n;i++)a[i]=I[n-i],b[i]=1ll*b[i]*J[i]%mod;
    
    while(lim<=2*n)lim<<=1,lo++;
    for(int i=1;i<lim;i++)r[i]=r[i>>1]>>1^(i&1)<<lo;
    NTT(a,1);NTT(b,1);
    for(int i=0;i<lim;i++)a[i]=1ll*a[i]*b[i]%mod;
    NTT(a,-1);
    for(int i=1;i<=n;i++)printf("%d\n",((1ll*n*C(n,i)-1ll*a[i+n]*I[i])%mod+mod)%mod);
    return 0;
}