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;
#define N 1111116
#define mod 924844033
#define next nxt

int n,x[N],y[N],next[N],p[N],s[N],r[N];
long long fac[N],inv[N],a[N],b[N];

void travel(int x){
	s[x]=1;
	for (int t=p[x];t;t=next[t])
		if (!s[y[t]])
		travel(y[t]),s[x]+=s[y[t]];
}

long long power(long long x,long long y){
	long long t=1;
	for (;y;x=x*x%mod,y/=2)
		if (y&1) t=t*x%mod;
	return t;
}

void dntt(long long a[],int k){
	int n=1<<19;
	for (int i=0;i<n;i++)
		if (r[i]>i) swap(a[i],a[r[i]]);
	for (int i=1;i<n;i*=2){
		long long t=power(5,mod-1+(mod-1)/(2*i)*k),p=1;
		for (int j=0;j<n;j++,p=p*t%mod)
			if ((i&j)==0){
				long long x=a[j],y=p*a[j+i]%mod;
				a[j]=(x+y)%mod,a[j+i]=(x-y)%mod;
			}
	}
}

void ntt(long long a[],long long b[]){
	int n=1<<19;
	for (int i=0;i<n;i++) r[i]=(r[i/2]/2)|((i&1)*n/2);
	dntt(a,1);dntt(b,1);
	for (int i=0;i<n;i++) a[i]=a[i]*b[i]%mod;
	dntt(a,-1);
	long long inv=power(n,mod-2);
	for (int i=0;i<n;i++) a[i]=a[i]*inv%mod;
}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();
	for (int i=1;i<=n-1;i++){
		x[2*i-1]=y[2*i]=read();
		y[2*i-1]=x[2*i]=read();
		next[2*i-1]=p[x[2*i-1]];p[x[2*i-1]]=2*i-1;
		next[2*i]=p[x[2*i]];p[x[2*i]]=2*i;
	}
	fac[0]=fac[1]=inv[0]=inv[1]=1;
	for (int i=2;i<=n;i++) fac[i]=fac[i-1]*i%mod;
	for (int i=2;i<=n;i++) inv[i]=-(mod/i)*inv[mod%i]%mod;
	for (int i=2;i<=n;i++) inv[i]=inv[i-1]*inv[i]%mod;
	travel(1);a[n]=n*fac[n]%mod;
	for (int i=2;i<=n;i++)
		a[s[i]]=(a[s[i]]-fac[s[i]])%mod,
		a[n-s[i]]=(a[n-s[i]]-fac[n-s[i]])%mod;
	for (int i=0;i<=n;i++) b[i]=inv[n-i];
	ntt(a,b);
	for (int k=1;k<=n;k++)
		printf("%lld\n",((a[n+k])%mod*inv[k]%mod+mod)%mod);
	return 0;
}