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 maxn = 5e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
vector<int> g[maxn];
ll dp[maxn], sum[maxn], fac[maxn], ifac[maxn], n;
ll fpow(ll a, ll b) {
    ll r = 1;
    while (b) {
        if (b & 1) r = r * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return r;
}
ll C(int x, int y) {
    if (x < 0 || x < y || y < 0) return 0;
    return fac[x] * ifac[y] % mod * ifac[x - y] % mod;
}
void dfs1(int u, int fa) {
    dp[u] = 1;
    for (auto it : g[u]) {
        if (it == fa) continue;
        dfs1(it, u);
        sum[u] += sum[it];
        dp[u] = dp[u] * dp[it] % mod * C(sum[u], sum[it]) % mod;
    }
    sum[u] += 1;
}
void dfs2(int u, int fa) {
    for (auto it : g[u]) {
        if (it == fa) continue;
        ll tmp = dp[u] * fpow(dp[it], mod - 2) % mod * fpow(C(n - 1, sum[it]), mod - 2) % mod;
        dp[it] = dp[it] * tmp % mod * C(n - 1, sum[it] - 1) % mod;
        dfs2(it, u);
    }
}
int main() {
    fac[0] = 1;
    for (int i = 1; i <= maxn - 10; i++) fac[i] = fac[i - 1] * i % mod;
    ifac[maxn - 10] = fpow(fac[maxn - 10], mod - 2);
    for (int i = maxn - 10 - 1; i >= 0; i--)
        ifac[i] = ifac[i + 1] * (i + 1) % mod;
    scanf("%lld", &n);
    for (int i = 1, x, y; i < n; i++) {
        scanf("%d%d", &x, &y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs1(1, 0); dfs2(1, 0);
    ll ans = 0;
    for (int i = 1; i <= n; i++)
    {
        printf("%lld\n", dp[i] % mod );
    }
    return 0;
}
