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>
using namespace std;
const int MAXN = 600005, P = 924844033, G = 5;
int n, head[MAXN], nxt[MAXN<<1], to[MAXN<<1], cnt;
inline void insert(int u, int v) { nxt[++cnt] = head[u]; to[cnt] = v; head[u] = cnt; }

int A[MAXN], B[MAXN], ans[MAXN], fac[MAXN], inv[MAXN], rev[MAXN], size[MAXN];
inline int QuickPow(int x, int k) {
	int res = 1;
	for(; k; k >>= 1, x = 1LL * x * x % P) if(k & 1) res = 1LL * res * x % P;
	return res;
}

inline void dfs(int u, int fa) {
	size[u] = 1;
	for(int v, i = head[u]; i; i = nxt[i]) if((v = to[i]) != fa) {
		dfs(v, u);
		size[u] += size[v];
		if(--A[size[v]] < 0) A[size[v]] += P;
	}
	if(--A[n-size[u]] < 0) A[n-size[u]] += P;
	if(++A[n] >= P) A[n] -= P;
}

inline void NTT(int* x, int len, int d) {
	for(int i = 0; i < len; ++i) rev[i] = rev[i>>1] >> 1 | ((i & 1) ? (len>>1) : 0);
	for(int i = 0; i < len; ++i) if(rev[i] > i) swap(x[i], x[rev[i]]);
	
	for(int h = 2; h <= len; h <<= 1) {
		int wn = QuickPow(G, (P - 1 + d * (P - 1) / h) % (P - 1));
		for(int j = 0; j < len; j += h) {
			int w = 1;
			for(int k = j; k < j + h / 2; ++k) {
				int u = x[k], v = 1LL * w * x[k+h/2] % P;
				x[k] = (u + v) % P;
				x[k+h/2] = (u - v + P) % P;
				w = 1LL * w * wn % P;
			}
		}
	}
	if(d > 0) return;
	int Inv = QuickPow(len, P - 2);
	for(int i = 0; i < len; ++i) x[i] = 1LL * x[i] * Inv % P;
}

inline void Mul() {
	int len = n << 1 | 1;
	while(len != (len&-len)) len += (len&-len);
	NTT(A, len, 1); NTT(B, len, 1);
	for(int i = 0; i < len; ++i) A[i] = 1LL * A[i] * B[i] % P;
	NTT(A, len, -1);
	for(int i = 1; i <= n; ++i) ans[i] = 1LL * inv[i] * A[n+i] % P;
}

int main() {
	scanf("%d", &n);
	for(int u, v, i = n; --i; ) {
		scanf("%d%d", &u, &v);
		insert(u, v);
		insert(v, u);
	}
	dfs(1, 0);
	
	fac[0] = fac[1] = inv[0] = inv[1] = 1;
	for(int i = 2; i <= n; ++i) {
		fac[i] = 1LL * i * fac[i-1] % P;
		inv[i] = 1LL * (P - P / i) * inv[P % i] % P;
	}
	for(int i = 2; i <= n; ++i) inv[i] = 1LL * inv[i] * inv[i-1] % P;
	for(int i = 0; i <= n; ++i) {
		A[i] = 1LL * A[i] * fac[i] % P;
		B[i] = inv[n-i];
	}
	
	Mul();
	
	for(int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
	return 0;
}