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>
#define ll long long
#define MAX 200005
#define MOD 1000000007
using namespace std;

ll modpow(ll a, ll b)
{
    if (b == 0) return 1;
    ll u = modpow(a, b / 2);
    u = (u * u)%MOD;
    if (b%2) u = (a * u)%MOD;
    return u;
}

int n;
vector<int> ad[MAX];
ll ans[MAX], sz[MAX];

void dfs(int s, int f)
{
    sz[s] = 1;
    for (auto nd:ad[s]) {
        if (nd != f) {
            dfs(nd, s);
            sz[s] += sz[nd];
        }
    }
    ans[1] = ans[1] * modpow(sz[s], MOD - 2)%MOD;
}

void dfs2(int s, int f)
{
    for (auto nd:ad[s]) {
        if (nd != f){
            ans[nd] = (ans[s] * sz[nd])%MOD * modpow(n - sz[nd], MOD - 2)%MOD;
            dfs2(nd, s);
        }
    }
}

int main()
{
    cin >> n;
    int a, b;
    ans[1] = 1;
    for (int i = 2; i <= n; i++) ans[1] = (ans[1]%MOD) * (i%MOD) % MOD;
    for (int i = 1; i <= n - 1; i++) {
        cin >> a >> b;
        ad[a].push_back(b);
        ad[b].push_back(a);
    }
    dfs(1, -1);
    dfs2(1, -1);
    for (int i = 1; i <= n; i++) {
        cout << ans[i] << "\n";
    }
}
