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<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define N 200010
#define mod 924844033
#define g 5
using namespace std;

struct edge{int x, y, next;}A[N*2];
int n, l, p[N], jc[N], inv[N], size[N], x, y, a[N*4], b[N*4], R[N*4], n1, L, INV, ans[N];

inline int pow(int y, int x){
	int s=1, t=y;
	while(x){if(x&1)s=(ll)s*t%mod; t=(ll)t*t%mod; x>>=1;}
	return s;
}

inline void add(int x, int y){A[++l].x=x; A[l].y=y; A[l].next=p[x]; p[x]=l;}
inline void dfs(int x, int fa){
	size[x]=1;
	for(int i=p[x]; i; i=A[i].next)if(fa!=A[i].y){dfs(A[i].y, x); size[x]+=size[A[i].y];}
}

inline int C(int n, int m){return (ll)jc[n]*inv[m]%mod*inv[n-m]%mod;}

inline void NTT(int *a, int f){
	for(int i=0; i<n1; i++)if(i<R[i])swap(a[i], a[R[i]]);
	for(int i=1; i<n1; i<<=1){
		int wn=pow(g, (mod-1)/(i*2)); if(f==-1)wn=pow(wn, mod-2);
		for(int j=0; j<n1; j+=i*2){
			int w=1;
			for(int k=0; k<i; k++){
				int x=a[j+k], y=(ll)w*a[j+k+i]%mod;
				a[j+k]=x+y; a[j+k+i]=x-y;
				if(a[j+k]>=mod)a[j+k]-=mod; if(a[j+k+i]<0)a[j+k+i]+=mod;
				w=(ll)w*wn%mod;
			}
		}
	}
	if(f==-1)for(int i=0; i<n1; i++)a[i]=(ll)a[i]*INV%mod;
}
inline void solve(){
	memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));
	for(int i=1; i<=n; i++)a[size[i]]++;
	for(int i=0; i<=n; i++)a[i]=(ll)a[i]*jc[i]%mod;
	for(int i=0; i<=n; i++)b[i]=inv[n-i];
	n1=n+1;
	L=1; while((1<<L)<n1*2)L++; n1=1<<L;
	for(int i=0; i<n1; i++)R[i]=(R[i/2]/2)|((i&1)<<(L-1));
	
	INV=pow(n1, mod-2);
	NTT(a, 1); NTT(b, 1);
	for(int i=0; i<n1; i++)a[i]=(ll)a[i]*b[i]%mod;
	NTT(a, -1);
	
	for(int i=1; i<=n; i++)ans[i]=(ans[i]-(ll)a[n+i]*inv[i]%mod+mod)%mod;
}

int main(){
	scanf("%d", &n);
	l=0; memset(p, 0, sizeof(p));
	for(int i=1; i<=n-1; i++){scanf("%d%d", &x, &y); add(x, y); add(y, x);}
	dfs(1, 0);
	jc[0]=1; for(int i=1; i<=n; i++)jc[i]=(ll)jc[i-1]*i%mod;
	inv[0]=inv[1]=1; for(int i=2; i<=n; i++)inv[i]=(ll)(mod-mod/i)*inv[mod%i]%mod;
	for(int i=1; i<=n; i++)inv[i]=(ll)inv[i-1]*inv[i]%mod;
	for(int i=1; i<=n; i++)ans[i]=(ll)(n+1)*C(n, i)%mod;
	solve();
	for(int i=1; i<=n; i++)size[i]=n-size[i];
	solve();
	for(int i=1; i<=n; i++)printf("%d\n", ans[i]);
	return 0;
}