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;
typedef long long ll;

inline void read(int&x){
	char c11=getchar();x=0;while(!isdigit(c11))c11=getchar();
	while(isdigit(c11))x=x*10+c11-'0',c11=getchar();
}

const int N = 601000, p = 924844033, G = 5;
struct Edge{int v,nxt;}a[N+N];
int head[N],sz[N],fac[N],inv[N];
int c[N];
int n,_;

inline int C(int nn,int mm){return (ll)fac[nn] * inv[mm]%p * inv[nn-mm]%p;}
inline int qm(const int&X){return X < p ? X : X - p;}
inline int qpow(int A,int B){
	int res = 1; while(B){
		if(B&1) res = (ll)res * A%p;
		A = (ll)A * A%p, B >>= 1;
	}return res;
}

inline void ad(){
	static int x,y; read(x), read(y);
	a[++_].v = y, a[_].nxt = head[x], head[x] = _;
	a[++_].v = x, a[_].nxt = head[y], head[y] = _;
}

void dfs(int x,int las){
	sz[x] = 1;
	for(int i=head[x];i;i=a[i].nxt)
		if(a[i].v != las){
			dfs(a[i].v,x);
			++c[sz[a[i].v]];
			sz[x] += sz[a[i].v];
		}
	++c[n-sz[x]];
}

namespace NTT{
	int rev[N];
	void ntt(int*a,int n,int sgn){
		for(int i=1;i<n;++i) if(i<rev[i]) swap(a[i], a[rev[i]]);
		for(int i=1;i<n;i<<=1){
			int gn = qpow(G,(p-1)/(i<<1));
			for(int j=0;j<n;j+=(i<<1)){
				int g = 1;
				for(int k=0;k<i;++k,g=(ll)g*gn%p){
					int x = a[j+k], y = (ll)g * a[j+k+i]%p;
					a[j+k] = qm(x+y), a[j+k+i] = qm(x-y+p);
				}
			}
		}
		if(!sgn){
			int iv = qpow(n,p-2); reverse(a+1,a+n);
			for(int i=0;i<n;++i) a[i] = (ll)iv * a[i]%p;
		}
	}
	void work(int*a,int*b,int n){
		int nn = 1, len = -1;
		while(nn <= n+n) nn <<= 1, ++len;
		for(int i=1;i<nn;++i) rev[i] = (rev[i>>1]>>1)|((i&1)<<len);
		ntt(a,nn,1), ntt(b,nn,1);
		for(int i=0;i<nn;++i) a[i] = (ll)a[i] * b[i]%p;
		ntt(a,nn,0);
	}
}

int arr[N],brr[N];

int main(){
	read(n);
	for(int i=1;i<n;++i) ad();
	dfs(1,0);
	
	fac[0] = inv[0] = 1;
	for(int i=1;i<=n;++i) fac[i] = (ll)fac[i-1] * i%p;
	inv[n] = qpow(fac[n], p-2);
	for(int i=n;i;--i) inv[i-1] = (ll)inv[i] * i%p;
	
	brr[0] = 1;
	for(int i=1;i<=n;++i){
		arr[i] = (ll)c[i] * fac[i]%p;
		brr[i] = inv[i];
	}
	reverse(arr,arr+n+1);
	NTT::work(arr,brr,n+1);
	
	for(int i=1;i<=n;++i)
		printf("%lld\n",((ll)C(n,i)*n - (ll)arr[n-i]*inv[i]%p+p)%p);
	return 0;
}