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;
vector<int> adj[200007];
ll sz[200007], dp[200007], f[200007];
int n;
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;
}
void init(){
    f[0]=1;
    for(int i=1;i<=200006;i++){
        f[i] = f[i-1]*i;
        f[i] %= MOD;
    }
}
void dfs1(int u, int p){
    sz[u]=1;
    for(auto v:adj[u]){
        if(v==p) continue;
        dfs1(v,u);
        sz[u]+=sz[v];
    }
    dp[1] = dp[1] * modpow(sz[u],MOD-2)%MOD;
}
void dfs2(int u, int p){
    for(auto v:adj[u]){
        if(v==p) continue;
        dp[v] = dp[u]%MOD * sz[v]%MOD * modpow(n-sz[v],MOD-2)%MOD;
        dfs2(v,u);
    }
}
int main()
{
    init();
    int a, b;
    cin >> n;
    dp[1] = f[n];
    for(int i=0;i<n-1;i++){
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    dfs1(1,-1);
    dfs2(1,-1);
    for(int i=1;i<=n;i++) cout << dp[i] << endl;
}
