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 LL long long
#define INF 0x7FFFFFFF//or 0x3f3f3f3f ? -Wall
using namespace std;

template<class T> inline
void read(T& x) {
	int f = 1; x = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9')   {if (ch == '-') f = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}

/*============ Header Template ============*/

const int N = 524288 + 5;
const int mo = 924844033;
const int g = 5;

int n, len;
int sz[N], rev[N];
vector<int> v[N];
LL c[N], a[N], b[N], f[N], wn[N];

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

LL get_pow(LL a, LL b) {
	LL ans = 1; a %= mo;
	while (b) {
		if (b & 1) ans = ans * a % mo;
		a = a * a % mo;
		b >>= 1;
	}
	return ans % mo;
}

LL getinv(LL x) {return get_pow(x, mo - 2);}

void NTT(LL *a, int len, int d = 1) {
	for (int i = 0; i < len; i++) if (i < rev[i]) swap(a[i], a[rev[i]]);
	int id = 0;
	for (int m = 2; m <= len; m <<= 1) {
		id++;
		LL wm = get_pow(g, (d * (mo - 1) / m) + mo - 1);
		for (int i = 0; i < len; 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 = w * wm % mo;
			}
		}
	}
	if (d == -1) {LL inv = getinv(len); for (int i = 0; i < len; i++) a[i] = a[i] * inv % mo;}
}

void pre() {
	for (int i = 1; i < len; i++) rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? (len >> 1) : 0);
}

int main() {
	read(n);
	for (int i = 1, x, y; i < n; i++) {
		read(x), read(y);
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(1, 0); c[n] = n; f[0] = 1;
	for (int i = 1; i <= n; i++) f[i] = f[i - 1] * i % mo;
	for (int i = 0; i <= n; i++) a[i] = c[i] * f[i] % mo;
	for (int i = 0; i <= n; i++) b[i] = getinv(f[i]);
	for (int i = 0; i <= n / 2; i++) swap(a[i], a[n - i]);
	for (len = 1; len < 2 * (n + 1); len <<= 1);
	pre(); NTT(a, len); NTT(b, len);
	for (int i = 0; i < len; i++) c[i] = a[i] * b[i] % mo;
	NTT(c, len, -1);
	for (int i = n - 1; i >= 0; i--) {
		LL ans = c[i];
		ans = ans * getinv(f[n - i]) % mo;
		printf("%lld\n", (ans + mo) % mo);
	}
	return 0;
}