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;
typedef long long ll;
#define rep(i,a,b) for(ll i=(a); i<(b); i++)
#define per(i,a,b) for(ll i=(a)-1; i>=(b); i--)
const ll mod=1e9+7, maxn=2e5+5;
ll n,a,b,f[maxn], sz[maxn], w[maxn], ans[maxn];
vector<ll> ady[maxn];
ll bin_pow(ll a, ll x) {
    ll ans=1;
    for(ll p=a;x;x/=2,p=p*p%mod) if(x&1) ans=ans*p%mod;
    return ans;
}
ll inv(ll a) {
    return bin_pow(a, mod-2);
}
void calc_fact() {
    f[0]=1;
    rep(i,1,maxn) f[i]=i*f[i-1]%mod;
}
ll choose(ll n, ll r) {
    return f[n]*inv(f[r])%mod*inv(f[n-r])%mod;
}
void dfs(ll x, ll p) {
    w[x]=sz[x]=1;
    for(ll to:ady[x]) if(to!=p) {
        dfs(to, x);
        sz[x]+=sz[to];
        w[x]=w[x]*choose(sz[x]-1,sz[to])%mod*w[to]%mod;
    }
}
void dfs2(ll x, ll p) {
    ans[x]=w[x];
    for(ll to: ady[x]) if(to!=p) {
        w[x]=w[x]*inv(choose(sz[x]-1, sz[to]))%mod*inv(w[to])%mod;
        sz[x]-=sz[to];
        sz[to]+=sz[x];
        w[to]=w[to]*choose(sz[to]-1,sz[x])%mod*w[x]%mod;
        dfs2(to,x);
        w[to]=w[to]*inv(choose(sz[to]-1, sz[x]))%mod*inv(w[x])%mod;
        sz[to]-=sz[x];
        sz[x]+=sz[to];
        w[x]=w[x]*choose(sz[x]-1,sz[to])%mod*w[to]%mod;
    }
}
int main() {
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    calc_fact();
    cin>>n;
    rep(i,1,n) cin>>a>>b, ady[a].push_back(b), ady[b].push_back(a);
    dfs(1,0);
    dfs2(1,0);
    rep(i,1,n+1) cout<<ans[i]<<"\n";
}
