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<iostream>
#include<cstring>
#define ll long long
using namespace std;
const int maxn = 2e5+5;
const int M = 1e9+7;
struct edge
{
    int u,v,nxt;
}es[maxn<<1];
int head[maxn],tot,n;
ll fac[maxn],inv[maxn],sz[maxn],dp[maxn],ans[maxn];
void add(int u,int v)
{
    es[++tot].u = u;
    es[tot].v = v;
    es[tot].nxt = head[u];
    head[u] = tot;
}
ll qpow(ll a,ll b)
{
    ll s = 1;
    while(b)
    {
        if(b&1) s = s*a%M;
        a = a*a%M;
        b>>=1;
    }
    return s;
}
void init()
{
    fac[0] = fac[1] =1;
    inv[0] = inv[1] = 1;
    for(int i=2;i<=n;i++) fac[i] = i*fac[i-1]%M;
    for(int i=2;i<=n;i++) inv[i] = qpow(fac[i],M-2);
}
ll C(ll a,ll b)
{
    ll res = 1ll*fac[a]*inv[b]%M*inv[a-b]%M;
    return res;
}
void dfs1(int u,int f)
{
    sz[u] = 1,dp[u]=1;
    for(int i=head[u];i!=-1;i=es[i].nxt)
    {
        int v = es[i].v;
        if(v==f) continue;
        dfs1(v,u);
        sz[u] += sz[v];
        dp[u] = 1ll*dp[u]*inv[sz[v]]%M*dp[v]%M;
    }
    dp[u] = 1ll*dp[u]*fac[sz[u]-1]%M;
}
void dfs2(int u,int f)
{
    if(u!=1) ans[u] = 1LL*ans[f]*qpow(C(n-1,sz[u]),M-2)%M*C(n-1,sz[u]-1)%M;
    for(int i=head[u];i!=-1;i=es[i].nxt)
    {
        int v = es[i].v;
        if(v==f) continue;
        dfs2(v,u);
    }
}
int main()
{
    memset(head,-1,sizeof(head));
    cin>>n;
    int u,v;
    for(int i=0;i<n-1;i++)
    {
        cin>>u>>v;
        add(u,v);
        add(v,u);
    }
    init();
    dfs1(1,-1);
    ans[1] = dp[1];
    dfs2(1,-1);
    for(int i=1;i<=n;i++) cout<<ans[i]<<endl;
    return 0;
}
