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 maxn 600005
#define rep(i,j,k) for(int i=(j),LIM=(k);i<=LIM;i++)
#define per(i,j,k) for(int i=(j),LIM=(k);i>=LIM;i--)
#define mod 924844033
using namespace std;

int n,sz[maxn],A[maxn],B[maxn];
vector<int>G[maxn];

void dfs(int u,int ff){
	sz[u] = 1;
	int v;
	rep(i,0,G[u].size()-1) if((v=G[u][i]) ^ ff){
		dfs(v,u);
		sz[u] += sz[v];
		A[sz[v]] -- , A[n-sz[v]]--;
	}
}

int Wl,Wl2,w[maxn],lg[maxn],inv[maxn],invf[maxn],fac[maxn];
int Pow(int b,int k){ int r=1;for(;k;k>>=1,b=1ll*b*b%mod) if(k&1) r=1ll*r*b%mod; return r; }
void init(int n){
	for(Wl=1;n>=Wl<<1;Wl<<=1);int pw=Pow(5,(mod-1)/(Wl2=Wl<<1));
	w[Wl] = inv[0] = inv[1] = fac[0] =fac[1] = invf[0] = invf[1] = 1;
	rep(i,Wl+1,Wl2) w[i] = 1ll * w[i-1] * pw % mod;
	per(i,Wl-1,1) w[i] = w[i<<1];
	rep(i,2,Wl2) lg [i] = lg[i>>1] + 1, inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod,
		fac[i] = 1ll * fac[i-1] * i % mod ,invf[i] = 1ll * invf[i-1] * inv[i] % mod;
}

int upd(int x){ return x += x >> 31 & mod; }
void NTT(int *A,int n,int tp){
	static int r[maxn]={};
	static unsigned long long ar[maxn];
	if(tp ^ 1) reverse(A+1,A+n);
	rep(i,0,n-1) r[i] = r[i>>1] >>1|(i&1)<<lg[n]-1 , ar[i] = upd(A[r[i]]);
 	for(int L=1;L<n;L<<=1) for(int s=0,L2=L<<1;s<n;s+=L2) for(int k=s,x=L,t;k<s+L;k++,x++)
 		t = ar[k+L] * w[x] % mod , ar[k+L] = ar[k] - t + mod , ar[k] += t;
 	rep(i,0,n-1) A[i] = ar[i] % mod;
	if(tp ^ 1) rep(i,0,n-1) A[i] = 1ll * A[i] * inv[n] % mod;
}

int main(){
	scanf("%d",&n);
	rep(i,1,n-1){
		int a,b;
		scanf("%d%d",&a,&b);
		G[a].push_back(b);
		G[b].push_back(a);
	}
	dfs(1,0);
	init(n << 1);
	rep(i,0,n) A[i] = 1ll * A[i] * fac[i] % mod , B[i] = invf[i];
	reverse(A,A+n+1);
	NTT(A,Wl2,1),NTT(B,Wl2,1);
	rep(i,0,Wl2-1) A[i] = 1ll * A[i] * B[i] % mod;
	NTT(A,Wl2,-1);
	rep(i,1,n){
		int ans = (1ll * fac[n] * invf[i] % mod * invf[n-i] % mod * n + A[n-i] * 1ll * invf[i]) % mod;
		printf("%d\n",(ans+mod)%mod);
	}
}