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 _for(i,a,b) for(int i = (a); i < (b); i++)
#define _rep(i,a,b) for(int i = (a); i <= (b); i++)
const int N = 2e5+100;
const int mod = 1e9+7;
int siz[N],n;
ll mul[N],fac[N],ans[N];
vector<int> g[N];

ll fpow(ll a, ll b) {
    ll ret = 1;
    while(b) {
        if(b&1)  ret = ret*a%mod;
        b >>= 1;
        a = a*a%mod;
    }
    return ret;
}
void predfs(int u, int fa) {
    siz[u] = mul[u] = 1;
    for(int v : g[u]) {
        if(v == fa) continue;
        predfs(v,u);
        siz[u] += siz[v];
        mul[u] = mul[u]*mul[v]%mod;
    }
    mul[u] = mul[u]*siz[u]%mod;
}
void dfs(int u, int fa, ll tmp) {
    for(int v : g[u]) if(v != fa) tmp = tmp*mul[v]%mod;
    ans[u] = fac[n-1]*fpow(tmp, mod-2)%mod;
    for(int v : g[u]) if(v != fa) dfs(v, u, tmp*fpow(mul[v], mod-2)%mod*(n-siz[v])%mod);
}
int main(){
    scanf("%d", &n);
    _for(i,1,n) {
        int x,y;         
        scanf("%d%d", &x, &y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    fac[0] = 1;
    _rep(i,1,n) fac[i] = fac[i-1]*i%mod;
    predfs(1, 0);
    dfs(1,0,1);
    _rep(i,1,n) printf("%lld\n", ans[i]);
    return 0;
}
