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 ll long long
#define ar array

const int mxN=2e5, M=1e9+7;
int n, s[mxN];
vector<int> adj[mxN];
ll iv[mxN], f1[mxN], f2[mxN], dp[mxN], ans[mxN];

void dfs1(int u=0, int p=-1) {
    dp[u]=1;
    for(int v : adj[u]) {
        if(v==p)
            continue;
        dfs1(v, u);
        dp[u]=dp[u]*dp[v]%M*f1[s[u]+s[v]]%M*f2[s[u]]%M*f2[s[v]]%M;
        s[u]+=s[v];
    }
    ++s[u];
}

ll mI(ll a, ll m) {
    return (a%=m)<2?a:(1-mI(m%a, a)*m)/a+m;
}

void dfs2(int u=0, int p=-1, ll pd=1) {
    ans[u]=dp[u]*pd%M*f1[n-1]%M*f2[s[u]-1]%M*f2[n-s[u]]%M;
    for(int v : adj[u]) {
        if(v==p)
            continue;
        dfs2(v, u, ans[u]*mI(dp[v], M)%M*f2[n-1]%M*f1[n-1-s[v]]%M*f1[s[v]]%M);
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> n;
    iv[1]=1;
    for(int i=2; i<n; ++i)
        iv[i]=M-M/i*iv[M%i]%M;
    f1[0]=f2[0]=1;
    for(int i=1; i<n; ++i) {
        f1[i]=f1[i-1]*i%M;
        f2[i]=f2[i-1]*iv[i]%M;
    }
    for(int i=1, u, v; i<n; ++i) {
        cin >> u >> v, --u, --v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    dfs1();
    dfs2();
    for(int i=0; i<n; ++i)
        cout << ans[i] << "\n";
}