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>
#define REP(i, a, b) for (int i = a; i <= b; ++i)
#define PER(i, a, b) for (int i = a; i >= b; --i)
using namespace std;

typedef long long LL;

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

const int N = 600005, mo = 924844033, g = 5;
int n, sz[N], rev[N];
LL a[N], b[N], f[N];
vector<int> to[N];

void dfs(int x, int f){
	sz[x] = 1;
	for (int i = 0; i < to[x].size(); ++i){
		int y = to[x][i];
		if (y == f) continue;
		dfs(y, x);
		sz[x] += sz[y];
		--a[sz[y]];
		--a[n - sz[y]];
	}
}

LL pwr(LL a, LL b){
	LL res = 1;
	for (; b; b >>= 1, (a *= a) %= mo)
		if (b & 1) (res *= a) %= mo;
	return res; 
}

LL inv(LL x){
	return pwr(x, mo - 2);
}

void ntt(LL *a, int n, int d = 1){
	REP(i, 0, n - 1) if (i > rev[i]) swap(a[i], a[rev[i]]);
	for (int m = 2; m <= n; m <<= 1){
		LL wn = pwr(g, d * (mo - 1) / m + mo - 1);
		for (int i = 0; i < n; i += m){
			LL w = 1;
			for (int j = i; j < i + m / 2; ++j){
				LL u = a[j], v = w * a[j + m / 2] % mo;
				a[j] = (u + v) % mo;
				a[j + m / 2] = (u - v) % mo;
				(w *= wn) %= mo;
			}
		}
	}
	if (d == -1){
		LL t = inv(n);
		REP(i, 0, n - 1) (a[i] *= t) %= mo;
	}
}

int main(){
	n = read();
	REP(i, 2, n){
		int u = read(), v = read();
		to[u].push_back(v);
		to[v].push_back(u);
	}
	dfs(1, 0); a[n] = n; f[0] = 1;
	REP(i, 1, n) f[i] = f[i - 1] * i % mo;
	REP(i, 0, n) a[i] = a[i] * f[i] % mo;
	REP(i, 0, n) b[i] = inv(f[i]);
	REP(i, 0, n / 2) swap(a[i], a[n - i]);
	int len;
	for (len = 1; len < 2 * (n + 1); len <<= 1);
	REP(i, 1, len - 1)
		if (i & 1) rev[i] = (rev[i >> 1] >> 1) ^ (len >> 1);
		else rev[i] = (rev[i >> 1] >> 1);
	ntt(a, len); ntt(b, len);
	REP(i, 0, len - 1) (a[i] *= b[i]) %= mo;
	ntt(a, len, -1);
	PER(i, n - 1, 0){
		LL ans = a[i] * inv(f[n - i]) % mo;
		printf("%lld\n", (ans + mo) % mo);
	}
	return 0;
}