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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=200010,Mo=924844033;
int n,tt;
int A[N*4],B[N*4],rev[N*4],head[N],to[N*2],nxt[N*2],sz[N],fac[N],inv[N];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline int Pow(int x,int y) {
    int ret=1;
    while(y) {
	if(y&1) ret=1LL*ret*x%Mo;
	x=1LL*x*x%Mo,y>>=1;
    }
    return ret;
}
inline void NTT(int *a,int n,int f) {
    for(int i=0;i<n;i++) if(i<rev[i]) swap(a[i],a[rev[i]]);
    for(int i=1;i<n;i<<=1) {
	int wn=Pow(5,(Mo-1)/(i<<1)),x,y;
	for(int j=0;j<n;j+=i<<1)
	    for(int k=0,w=1;k<i;k++,w=1LL*w*wn%Mo) {
		x=a[j+k],y=1LL*a[j+k+i]*w%Mo;
		a[j+k]=(x+y)%Mo,a[j+k+i]=(x-y)%Mo;
	    }
    }
    if(f==1) return;
    reverse(a+1,a+n);
    int Inv=Pow(n,Mo-2);
    for(int i=0;i<n;i++) a[i]=1LL*a[i]*Inv%Mo;
}
inline void dfs(int x) {
    sz[x]=1;
    for(int i=head[x];i;i=nxt[i]) {
	int y=to[i];
	if(!sz[y]) {
	    dfs(y),sz[x]+=sz[y];
	    --A[sz[y]],--A[n-sz[y]];
	}
    }
}
int main() {
    int s,L;
    cin>>n,A[n]=n;
    for(int i=1;i<n;i++) {
	int x=gi(),y=gi();
	to[++tt]=y,nxt[tt]=head[x],head[x]=tt;
	to[++tt]=x,nxt[tt]=head[y],head[y]=tt;
    }
    dfs(1),fac[0]=1;
    for(int i=1;i<=n;i++) fac[i]=1LL*fac[i-1]*i%Mo;
    inv[n]=Pow(fac[n],Mo-2);
    for(int i=n;i;i--) inv[i-1]=1LL*inv[i]*i%Mo;
    for(int i=1;i<=n;i++) A[i]=1LL*A[i]*fac[i]%Mo;
    for(int i=0;i<=n;i++) B[i]=inv[n-i];
    for(s=1,L=0;s<=n+n;s<<=1) L++;
    for(int i=0;i<s;i++) rev[i]=((rev[i>>1]>>1)|((i&1)<<L-1));
    NTT(A,s,1),NTT(B,s,1);
    for(int i=0;i<s;i++) A[i]=1LL*A[i]*B[i]%Mo;
    NTT(A,s,-1);
    for(int i=1;i<=n;i++) printf("%lld\n",1LL*(A[n+i]+Mo)*inv[i]%Mo);
    return 0;
}