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<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;

typedef long long ll;
const int MAXN=200005*4;
const int MOD=924844033;
const int G=5;

int n;
int siz[MAXN],cnt[MAXN];
ll a[MAXN],b[MAXN];
vector<int> E[MAXN];
ll fac[MAXN],finv[MAXN];

ll ksm(ll x,ll y){
    ll res=1;
    while(y){
        if(y&1) res=res*x%MOD;
        x=x*x%MOD;
        y>>=1;
    }
    return res;
}

void Init(){
    fac[0]=1;
    for(int i=1;i<MAXN;i++) fac[i]=fac[i-1]*i%MOD;
    finv[1]=1;
    for(int i=2;i<MAXN;i++) finv[i]=1ll*(MOD-MOD/i)*finv[MOD%i]%MOD;
    finv[0]=1;
    for(int i=1;i<MAXN;i++) finv[i]=finv[i]*finv[i-1]%MOD;
}

void dfs(int u,int fa=0){
    siz[u]=1;
    for(int i=0;i<(int)E[u].size();i++){
        int v=E[u][i];
        if(v==fa) continue;
        dfs(v,u);
        siz[u]+=siz[v];
    }
    cnt[siz[u]]--;
    cnt[n-siz[u]]--;
}

void ntt(ll A[],int n,int mode){
    for(int i=1,j=0;i<n;i++){
        for(int s=n;j^=s>>=1,~j&s;);
        if(i<j) swap(A[i],A[j]);
    }
    for(int i=1;i<n;i<<=1){
        ll w1=ksm(G,(MOD-1)/(i<<1)),w=1;
        if(mode==-1) w1=ksm(w1,MOD-2);
        for(int j=0;j<i;j++,w=(1ll*w*w1)%MOD)
            for(int l=j,r=l+i;l<n;l+=(i<<1),r=l+i){
                int temp=1ll*A[r]*w%MOD;
                A[r]=(A[l]-temp+MOD)%MOD;
                A[l]=(A[l]+temp)%MOD;
            }
    }
    if(mode==-1){
        ll inv=ksm(n,MOD-2);
        for(int i=0;i<n;i++)
            A[i]=1ll*inv*A[i]%MOD;
    }
}

int main(){

    Init();
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        E[u].push_back(v);
        E[v].push_back(u);
    }
    dfs(1);
    cnt[n]=n;
    for(int i=1;i<=n;i++){
        a[i]=cnt[i]*fac[i]%MOD;
        b[i]=finv[n-i];
        //printf("%lld %lld\n",a[i],b[i]);
    }
    ll len=1;
    for(;len<=n*2;len<<=1);
    ntt(a,len,1),ntt(b,len,1);
    for(int i=0;i<=len;i++)
        a[i]=a[i]*b[i]%MOD;
    ntt(a,len,-1);
    for(int i=1;i<=n;i++){
        ll ans=a[n+i]*finv[i]%MOD;
        ans=(ans+MOD)%MOD;
        printf("%lld\n",ans);
    }

    fclose(stdin);
    fclose(stdout);
}