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;


using ll = long long;
const int Mod = 1e9+7;
const int MaxN = 2e5+10;
int N;
ll fac[MaxN], facinv[MaxN];
vector<int> adj[MaxN];
int sz[MaxN];
ll dp[MaxN], num[MaxN];


void exgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0) {
        x = 1, y = 0;
        return;
    }
    exgcd(b, a % b, y, x);
    y -= a / b * x;
}


ll invmod(ll a, ll m)
{
    ll x, y;
    exgcd(a, m, x, y);
    return (x % m + m) % m;
}


void init()
{
    fac[0] = 1;
    for (int i = 1; i < MaxN; i++)
        fac[i] = fac[i - 1] * i % Mod;
    facinv[MaxN - 1] = invmod(fac[MaxN - 1], Mod);
    for (int i = MaxN - 2; i >= 0; i--)
        facinv[i] = facinv[i + 1] * (i + 1) % Mod;
}


void dfs_dp(int u, int fa)
{
    int sz_u = 0;
    ll dp_u = 1;
    for (int v : adj[u]) {
        if (v == fa) continue;
        dfs_dp(v, u);
        sz_u += sz[v];
        dp_u = dp_u * dp[v] % Mod * facinv[sz[v]] % Mod;
    }
    sz[u] = sz_u + 1;
    dp[u] = dp_u * fac[sz_u] % Mod;
}


void dfs_num(int u, int fa)
{
    num[u] = (u == 1) ? dp[u] : (num[fa] * sz[u] % Mod * invmod(N - sz[u], Mod) % Mod);
    for (int v : adj[u]) {
        if (v != fa) dfs_num(v, u);
    }
}


void solve()
{
    dfs_dp(1, 0);
    dfs_num(1, 0);
    for (int i = 1; i <= N; i++) cout << num[i] << endl;
}


int main()
{
    ios::sync_with_stdio(false); cin.tie(0);

    cin >> N;
    int a, b;
    for (int i = 1; i <= N - 1; i++) {
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    init();
    solve();
    return 0;
}
