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>
#define mod 924844033
using namespace std;
int n,x,y;
int fir[200005],nxt[400005],to[400005],cnt;
int siz[200005],Cnt[200005];
int fac[200005],ifac[200005];
int lim=1,L=0,f[800005],g[800005],R[800005],Inv;

int read(){
    int x=0;char ch=getchar();
    while(!isdigit(ch)) ch=getchar();
    while(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
    return x;
}

void print(int x){
    if(x>=10) print(x/10);
    putchar(x%10+'0');
}

void dfs(int x,int f){
    siz[x]=1;
    for(int i=fir[x];i;i=nxt[i]){
        if(to[i]==f) continue;
        dfs(to[i],x);
        siz[x]+=siz[to[i]];
        Cnt[siz[to[i]]]++;
    }
    if(x!=1) Cnt[n-siz[x]]++;
}

int ksm(int x,int y){
    int res=1;
    while(y){
        if(y&1) res=1ll*res*x%mod;
        x=1ll*x*x%mod,y/=2;
    }
    return res;
}

int C(int x,int y){
    int res=1ll*fac[x]*ifac[y]%mod*ifac[x-y]%mod;
    return res;
}

void NTT(int *x,int on){
    for(int i=0;i<lim;i++) if(i<R[i]) swap(x[i],x[R[i]]);
    for(int i=2;i<=lim;i*=2){
        int wn=ksm(5,(mod-1)/i);
        if(on==-1) wn=ksm(wn,mod-2);
        for(int j=0;j<lim;j+=i){
            int w=1;
            for(int k=0;k<i/2;k++){
                int u=x[j+k],v=1ll*w*x[j+k+i/2]%mod;
                x[j+k]=(u+v)%mod;
                x[j+k+i/2]=(u+mod-v)%mod;
                w=1ll*w*wn%mod;
            }
        }
    }
}

int main(){
    n=read();
    fac[0]=1;for(int i=1;i<=n;i++) fac[i]=1ll*i*fac[i-1]%mod;
    ifac[n]=ksm(fac[n],mod-2);
    for(int i=n-1;i>=0;i--) ifac[i]=1ll*(i+1)*ifac[i+1]%mod;
    for(int i=1;i<n;i++){
        x=read(),y=read();
        to[++cnt]=y,nxt[cnt]=fir[x],fir[x]=cnt;
        to[++cnt]=x,nxt[cnt]=fir[y],fir[y]=cnt;
    }
    dfs(1,0);
    while(lim<=2*n) lim*=2,L++;
    for(int i=0;i<lim;i++) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
    for(int i=1;i<=n;i++) f[i]=1ll*Cnt[i]*fac[i]%mod;
    for(int i=0;i<=n;i++) g[i]=ifac[n-i];
    NTT(f,1),NTT(g,1);
    for(int i=0;i<lim;i++) f[i]=1ll*f[i]*g[i]%mod;
    NTT(f,-1);
    Inv=ksm(lim,mod-2);
    for(int i=1,v;i<=n;i++){
        v=(1ll*n*C(n,i)%mod+mod-1ll*ifac[i]*f[n+i]%mod*Inv%mod)%mod;
        print(v),puts("");
    }

    return 0;
}