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 pb push_back
#define ll unsigned long long
#define maxn 500005
#define fr(i,j,k) for(int i=j;i<k;i++)
#define f(n) fr(i,0,n)
#define f1(n) fr(i,1,n+1)
#define ms(i) memset(i,0,sizeof(i));
#define ms1(i) memset(i,-1,sizeof(i));
#define F first
#define S second
#define all(x) x.begin(), x.end()
const int mod = 1e9 + 7;
vector<int>g[200005];
ll ans[200005];
ll sz[200005];
ll pre[maxn];
ll inv[maxn];
ll prei[maxn];
int n;
void build(int n){
    pre[1] = pre[0] = 1, inv[1] = inv[0] = 1, prei[1] = prei[0] = 1;
    for(int i = 2 ; i <= n ; i++){
        pre[i] = pre[i-1] * i % mod;
        inv[i] = (mod - mod / i * inv[mod % i] % mod) % mod;
        prei[i] = prei[i-1] * inv[i] % mod;
    }
}
void dfs(int now, int pre) {
    sz[now] = 1;
    for (auto i:g[now]) {
        if (i == pre) continue;
        dfs(i,now);
        sz[now] += sz[i];
    }
}
void dfs2(int now,int pre, ll a) {
    ans[now] = a;
    for (auto i:g[now]) {
        if (i == pre)continue;
        ll ca = a;
        ca *= sz[i];
        ca %= mod;
        ca *= inv[n-sz[i]];
        ca %= mod;
        dfs2(i,now,ca);
    }
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    f(n - 1) {
        int x ,y;
        cin >> x >> y;
        g[x].pb(y);
        g[y].pb(x);
    }
    dfs(1,1);
    build(n);
    ll a = 1;
    f1(n) {
        a *= i;
        a %= mod;
    }
    f1(n) {
        a *= inv[sz[i]];
        a %= mod;
    }
    dfs2(1,1,a);
    f1(n) {
        cout << ans[i] << '\n';
    }
}
