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 N = 2e5 + 6, MOD = 1e9 + 7;
ll f[N], fi[N];
ll c(ll n, ll k)
{
    return f[n] * fi[k] % MOD * fi[n - k] % MOD;
}
ll inv(ll x)
{
    ll a = x, b = MOD - 2, r = 1;
    for (; b; b >>= 1, a = a * a % MOD)
        if (b & 1)
            r = r * a % MOD;
    return r;
}
int n, sz[N];
vector<int> g[N];
ll dp[N], ans[N];
void dfs(int u, int p)
{
    sz[u] = 1;
    dp[u] = 1;
    for (int v : g[u])
        if (v != p) {
            dfs(v, u);
            sz[u] += sz[v];
            dp[u] = dp[u] * c(sz[u] - 1, sz[v]) % MOD * dp[v] % MOD;
        }
}
void go(int u, int p, ll pw = 1)
{
    ans[u] = dp[u];
    if (u != 1)
        ans[u] = ans[u] * c(n - 1, n - sz[u]) % MOD * pw % MOD;
    for (int v : g[u])
        if (v != p) {
            go(v, u, ans[u] * inv(dp[v] * c(n - 1, sz[v]) % MOD) % MOD);
        }
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    f[0] = f[1] = fi[1] = fi[0] = 1;
    for (int i = 2; i < N; ++i)
        f[i] = i * f[i - 1] % MOD;
    for (int i = 2; i < N; ++i)
        fi[i] = (MOD - MOD / i) * fi[MOD % i] % MOD;
    for (int i = 2; i < N; ++i)
        fi[i] = fi[i] * fi[i - 1] % MOD;
    cin >> n;
    for (int i = 0; i < n - 1; ++i) {
        int a, b;
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    dfs(1, 1);
    go(1, 1);
    for (int i = 1; i <= n; ++i)
        cout << ans[i] << '\n';
}
