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>
typedef long long ll;
using namespace std;

const int N = 2e5 + 10;
const ll mod = 1e9 + 7;
int n, a, b;
vector<int>v[N];

ll ksm(ll a, ll p)
{
    ll ans = 1;
    while (p)
    {
        if (p & 1) ans = ans * a % mod;
        a = a * a % mod;
        p >>= 1;
    }
    return ans;
}

ll dp[N], siz[N];
void dfs1(int son, int fa)
{
    siz[son] = 1;
    for (auto x: v[son])
    {
        if (x == fa) continue;
        dfs1(x, son);
        siz[son] += siz[x];
    }
}
void dfs2(int son, int fa)
{
    for (auto x: v[son])
    {
        if (x == fa) continue;
        dp[x] = dp[son] * ksm(siz[x], mod - 2) % mod * (n - siz[x]) % mod;
        dfs2(x, son);
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i < n; i++)
    {
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    
    dfs1(1, 0);
    dp[1] = 1;
    for (int i = 1; i <= n; i++) dp[1] = dp[1] * siz[i] % mod;
    dfs2(1, 0);
    
    ll ans = 1;
    for (int i = 1; i <= n; i++) ans = ans * i % mod;
    for (int i = 1; i <= n; i++) cout << ans * ksm(dp[i], mod - 2) % mod << endl;
}