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;
const int INF = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const int N = 200005;

vector<int> G[N];
int sz[N];
LL ans[N];
LL now = 1;
LL pow_mod(LL x, LL n) {
    LL res = 1;
    while (n) {
        if (n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}
void dfs1(int u, int fa) {
    sz[u] = 1;
    for (auto v : G[u]) {
        if (v == fa) continue;
        dfs1(v, u);
        sz[u] += sz[v];
    }
}
int n;
void dfs2(int u, int fa) {
    ans[u] = pow_mod(now, mod - 2);
    for (auto v : G[u]) {
        if (v == fa) continue;
        now = now * pow_mod(sz[v], mod - 2) % mod * (n - sz[v]) % mod;
        dfs2(v, u);
        now = now * pow_mod(n - sz[v], mod - 2) % mod * sz[v] % mod;
    }
}
int main() {
    cin >> n;
    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs1(1, 0);
    for (int i = 1; i <= n; i++) {
        now = now * sz[i] % mod;
    }
    dfs2(1, 0);
    LL wc = 1;
    for (int i = 1; i <= n; i++) wc = wc * i % mod;
    for (int i = 1; i <= n; i++) {
        printf("%lld\n", ans[i] * wc % mod);
    }
    return 0;
}
