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;

const int N = 200000 + 5, mod = 1e9 + 7;

int fpow (int base, int v) {
    int tot = 1;
    for (; v; v >>= 1, base = 1LL * base * base % mod) if (v & 1) tot = 1LL * tot * base % mod;
    return tot;
}

vector<int> G[N];
int n, dp[N], sz[N], fac[N], ifac[N], ans[N];

void dfs (int u, int v) {
    dp[u] = 1, sz[u] = 1;
    for (int i = 0; i < (signed)G[u].size(); ++i) if (G[u][i] != v) {
        dfs (G[u][i], u);
        sz[u] += sz[G[u][i]];
        dp[u] = 1LL * dp[u] * dp[G[u][i]] % mod * ifac[sz[G[u][i]]] % mod;
    }
    dp[u] = 1LL * dp[u] * fac[sz[u] - 1] % mod;
}

void dfs2 (int u, int v) {
    if (u != 1)
        ans[u] = 1LL * dp[u] * ifac[sz[u] - 1] % mod * fac[n - 1] % mod * v % mod * ifac[n - sz[u]] % mod;
    int tmp = 1LL * dp[u] * ifac[sz[u] - 1] % mod * v % mod * ifac[n - sz[u]] % mod;
    for (int i = 0; i < (signed)G[u].size(); ++i) if (sz[G[u][i]] < sz[u]) {
        dfs2 (G[u][i], 1LL * tmp * fac[n - sz[G[u][i]] - 1] % mod * fpow (dp[G[u][i]], mod - 2) % mod * fac[sz[G[u][i]]] % mod);
    }
}

int main()
{
    scanf("%d", &n);

    fac[0] = 1; for (int i = 1; i <= n; ++i) fac[i] = 1LL * fac[i - 1] * i % mod;
    ifac[n] = fpow (fac[n], mod - 2);
    for (int i = n - 1; i >= 0; --i) ifac[i] = 1LL * ifac[i + 1] * (i + 1) % mod;

    for (int i = 1; i < n; ++i) {
        int a, b; scanf("%d%d", &a, &b);
        G[a].push_back (b), G[b].push_back (a);
    }
    dfs (1, 0);
    ans[1] = dp[1];
    dfs2 (1, 1);
    for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
    return 0;
}