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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

typedef vector<int> vi;

const int Z = 1e9+7;

void dfs1(vi e[], int inv[], int x, int p[], int sz[], int dp[]) {
    sz[x] = 1;
    dp[x] = 1;

    for (int y : e[x]) {
        if (y != p[x]) {
            p[y] = x;
            dfs1(e, inv, y, p, sz, dp);
            sz[x] += sz[y];
            dp[x] = 1ll * dp[x] * dp[y] % Z;
        }
    }

    dp[x] = 1ll * dp[x] * inv[sz[x]] % Z;
}

void dfs2(vi e[], int inv[], int x, int p[], int sz[], int dp[]) {
    dp[x] = x == 0 ? dp[x] : 1ll * dp[p[x]] * sz[x] % Z * inv[sz[0]-sz[x]] % Z;

    for (int y : e[x]) {
        if (y != p[x]) {
            dfs2(e, inv, y, p, sz, dp);
        }
    }
}

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

    int n;
    cin >> n;
    vi e[n];
    for (int i = 0; i < n - 1; i++) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        e[x].push_back(y);
        e[y].push_back(x);
    }

    int inv[n+1];
    for (int i = 1; i <= n; i++) {
        inv[i] = i == 1 ? 1 : 1ll * (Z - Z / i) * inv[Z%i] % Z;
    }

    int p[n], sz[n], dp[n];
    p[0] = -1;
    dfs1(e, inv, 0, p, sz, dp);

    dfs2(e, inv, 0, p, sz, dp);

    int facn = 1;
    for (int i = 1; i <= n; i++) {
        facn = 1ll * facn * i % Z;
    }

    for (int i = 0; i < n; i++) {
        cout << 1ll * facn * dp[i] % Z << endl;
    }

    return 0;
}
