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 mod ((int)1e9+7)
using namespace std;
typedef pair<int,int> pi;
typedef long long int ll;

const int N = (int)2e5+5;
vector<int> v[N];
ll power(ll a,ll b) {
    if (b==0) return 1;
    ll c = power(a,b>>1);
    c = (c*c)%mod;
    if (b&1) return (c*a)%mod;
    return c;
}
ll f[N], dp[N], ans[N];
ll comb(int n,int r) {
    ll num = f[n];
    ll den = (f[r]*f[n-r])%mod;
    return (num*power(den,mod-2))%mod;
}
int cnt[N];
void dfs(int i,int p) {
    cnt[i] = 1;
    for(auto it:v[i]) {
        if(it!=p) {
            dfs(it,i);
            cnt[i] += cnt[it];
        }
    }
    int n = cnt[i]-1;
    dp[i] = 1;
    for(auto it:v[i]) {
        if(it!=p) {
            dp[i] *= (comb(n,cnt[it])*dp[it])%mod;
            dp[i] %= mod;
            n -= cnt[it];
        }
    }
}
int n;
void run(int i,int p) {
    ans[i] = dp[i];
    for(auto it:v[i]) {
        if(it!=p) {
            int t = (comb(n-1,cnt[it])*dp[it])%mod;
            dp[i] = (dp[i]*power(t,mod-2))%mod;
            int s = (comb(n-1,n-cnt[it])*dp[i])%mod;
            dp[it] = (dp[it]*s)%mod;
            run(it,i);
            dp[it] = (dp[it]*power(s,mod-2))%mod;
            dp[i] = (dp[i]*t)%mod;
        }
    }
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    f[0] = 1;
    for(int i=1;i<N;i++) {
        f[i] = (f[i-1]*i)%mod;
    }
    cin>>n;
    for(int i=1,x,y;i<n;i++) {
        cin>>x>>y;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    dfs(1,1);
    run(1,1);
    for(int i=1;i<=n;i++) {
        cout<<ans[i]<<"\n";
    }
}
