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
const ll N = 2e5 + 5, mod = 1e9 + 7;
vector<ll> g[N];
ll a[N], n, sz[N], tmp = 1;

ll ksm(ll a, ll n) {
    ll ans = 1;
    while (n) {
        if (n & 1) ans = ans * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ans;
}

void dfs(int u, int fa) {
    sz[u] = 1;
    for (auto v : g[u]) {
        if (v != fa) {
            dfs(v, u);
            sz[u] += sz[v];
        }
    }
    tmp = tmp * sz[u] % mod;
}

void fun(int u, int fa) {
    for (auto v : g[u]) {
        if (v != fa) {
            a[v] = a[u] * ksm(sz[v], mod - 2) % mod * (n - sz[v]) % mod;
            fun(v, u);
        }
    }
}

int main() {
    cin >> n;
    ll jc = 1;
    for (ll i= 2, u, v; i <= n; ++i) {
        cin >> u >> v;
        g[u].push_back(v), g[v].push_back(u);
        jc = jc * i % mod;
    }
    dfs(1, 0);
    a[1] = tmp;
    fun(1, 0);
    for (int i = 1; i <= n; ++i) {
        printf("%lld\n", jc * ksm(a[i], mod - 2) % mod);
    }
    return 0;
}