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;
#define cs const 
#define int long long
#define low(i) (i&(-i))
#define mid ((l+r)>>1)
#define ls (now<<1)
#define rs (now<<1|1)
#define par pair<int,int>
#define mp make_pair
#define fr first 
#define se second
#define rep(i, x, y) for(int i=x; i<=y; ++i)
#define drep(i, x, y) for(int i=x; i>=y; --i)
#define fore(i, x) for(int i=fir[x]; i; i=nex[i])
cs int base = 233;
cs int ff = 2e6+1;
cs int mod = 1e9+7;
cs int inf = 1e9+1;
int n, x, y, ans[ff];
int fir[ff], nex[ff], to[ff], tot;
void add(int x, int y)
{
	nex[++tot] = fir[x];
	fir[x] = tot;
	to[tot] = y;
}
int ksm(int a, int b, int ha = 1) 
{
    while (b) 
	{
        if (b & 1)
            ha = ha * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return ha;
}
int inv(int x) { return ksm(x, mod-2) % mod; }
int siz[ff], f[ff], u, v;
void dfs1(int now, int las)
{
	siz[now] = 1;
	fore(i, now)
		if( to[i] != las )
			dfs1(to[i], now), siz[now] += siz[to[i]];
	f[1] = ( f[1] * siz[now] ) % mod;
}
void dfs2(int now, int las)
{
	if( now != 1 ) 
		f[now] = f[las] * ( n - siz[now] ) % mod * inv(siz[now]) % mod;
	fore(i, now)
		if( to[i] != las )
			dfs2(to[i], now);
}
void init()
{
	cin >> n;
	int ps = 1;
	rep(i, 1, n-1)
		scanf("%lld %lld", &u, &v), add(u, v), add(v, u), ps = ps * i % mod;
	ps = ps * n % mod;
	f[1] = 1; dfs1(1, 0); dfs2(1, 0);
	rep(i, 1, n)
		cout << ( inv(f[i]) * ps ) % mod << endl;
}
signed main()
{
	int T = 1;
	while(T--) 
		init();
} 