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;
const int mod = 1e9+7;
typedef long long ll;
vector<int> g[200002];
ll dp[200002];
int sz[200002];
ll ans[200002];
ll f1[200002], f2[200002];
ll pow2(ll a, ll b){
    ll ans = 1;
    while(b){
        if(b%2 == 1) ans = ans*a%mod;
        a = a*a%mod;
        b/=2;
    }
    return ans;
}
ll ncr(int a, int b){
    return f1[a]*f2[b]%mod*f2[a-b]%mod;
}
void dfs1(int u, int p){
    sz[u] = 1;
    dp[u] = 1;
    for(int i=0;i<g[u].size();i++){
        int v = g[u][i];
        if(v == p) continue;
        dfs1(v, u);
        sz[u] += sz[v];
        dp[u] = dp[u]*dp[v]%mod;
        dp[u] = dp[u]*f2[sz[v]]%mod;
    }
    dp[u] = dp[u]*f1[sz[u]-1]%mod;
}
int n;
void dfs2(int u, int p, int sz2, ll pv){
    ll res = pv, comb = f2[sz2];
    for(int i=0;i<g[u].size();i++){
        int v = g[u][i];
        if(v == p) continue;
        res = res*dp[v]%mod;
        comb = comb*f2[sz[v]]%mod;
        sz2 += sz[v];
    }
    ans[u] = res*f1[n-1]%mod*comb%mod;
    for(int i=0;i<g[u].size();i++){
        int v = g[u][i];
        if(v == p) continue;
        dfs2(v, u, sz2-sz[v]+1, res*pow2(dp[v], mod-2)%mod*f1[sz2-sz[v]]%mod*comb%mod*pow2(f2[sz[v]], mod-2)%mod);
    }
}
int main(){
    f1[0] = f2[0] = 1;
    for(int i=1;i<=200000;i++) f1[i] = f1[i-1]*i%mod, f2[i] = pow2(f1[i], mod-2);
    scanf("%d", &n);
    for(int i=1;i<=n-1;i++){
        int a, b;
        scanf("%d%d", &a, &b);
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }
    dfs1(1, 0);
    dfs2(1, 0, 0, 1);
    for(int i=1;i<=n;i++){
        printf("%lld\n", ans[i]);
    }
}