We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and b_i. For each k=1, ..., N, solve the problem below:

* Consider writing a number on each vertex in the tree in the following manner:
* First, write 1 on Vertex k.
* Then, for each of the numbers 2, ..., N in this order, write the number on the vertex chosen as follows:
* Choose a vertex that still does not have a number written on it and is adjacent to a vertex with a number already written on it. If there are multiple such vertices, choose one of them at random.
* Find the number of ways in which we can write the numbers on the vertices, modulo (10^9+7).

Constraints

* 2 \leq N \leq 2 \times 10^5
* 1 \leq a_i,b_i \leq N
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}


Output

For each k=1, 2, ..., N in this order, print a line containing the answer to the problem.

Examples

Input

3
1 2
1 3


Output

2
1
1


Input

2
1 2


Output

1
1


Input

5
1 2
2 3
3 4
3 5


Output

2
8
12
3
3


Input

8
1 2
2 3
3 4
3 5
3 6
6 7
6 8


Output

40
280
840
120
120
504
72
72
#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
vector<int> g[maxn];
ll dp[maxn],sum[maxn],fac[maxn],ifac[maxn],n;
ll fpow(ll a,ll b) {
	ll r = 1;
	while (b) {
		if (b & 1) r = r * a % mod;
		b >>= 1;
		a = a * a % mod;
	}
	return r;
}
ll C(int x,int y) {
	if (x < 0 || x < y || y < 0) return 0;
	return fac[x] * ifac[y] % mod * ifac[x - y] % mod;
}
void dfs1(int u,int fa) {
	dp[u] = 1;
	for (auto it : g[u]) {
		if (it == fa) continue;
		dfs1(it,u);
		sum[u] += sum[it];
		dp[u] = dp[u] * dp[it] % mod * C(sum[u],sum[it]) % mod;
	}
	sum[u] += 1;
}
void dfs2(int u,int fa) {
	for (auto it : g[u]) {
		if (it == fa) continue;
		ll tmp = dp[u] * fpow(dp[it],mod - 2) % mod * fpow(C(n - 1,sum[it]),mod - 2) % mod;
		dp[it] = dp[it] * tmp % mod * C(n - 1,sum[it] - 1) % mod;
		dfs2(it,u);
	}
}
int main() {
	fac[0] = 1;
	for (int i = 1; i <= maxn - 10; i++) fac[i] = fac[i - 1] * i % mod;
	ifac[maxn - 10] = fpow(fac[maxn - 10],mod - 2);
	for (int i = maxn - 10 - 1; i >= 0; i--)
		ifac[i] = ifac[i + 1] * (i + 1) % mod;
	scanf("%lld",&n);
	for (int i = 1,x,y; i < n; i++) {
		scanf("%d%d",&x,&y);
		g[x].push_back(y);
		g[y].push_back(x);
	}
	dfs1(1,0); dfs2(1,0);
	for (int i = 1; i <= n; i++)
		printf("%lld\n",dp[i]);
	return 0;
}