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 sz = 4e5+5, mod = 1e9+7;

ll n, dp[sz], cnt[sz], ans[sz];
vector <int> g[sz];

ll fac[sz], finv[sz], inv[sz];

void prec()
{
    inv[0] = inv[1] = 1;
    for(int i=2; i<sz; i++)
        inv[i] = mod - mod / i * inv[mod % i] % mod;
    fac[0] = finv[0] = 1;
    for(int i=1; i<sz; i++) {
        fac[i] = fac[i-1] * i % mod;
        finv[i] = finv[i-1] * inv[i] % mod;
    }
}

ll cmb(int n, int r)
{
    ll d = finv[r] * finv[n-r] % mod;
    return fac[n] * d % mod;
}

ll invmod(ll b)
{
    ll ans = 1, n = mod - 2;
    while(n) {
        if(n&1) ans = (ans * b) % mod;
        n >>= 1;
        b = (b * b) % mod;
    }
    return ans;
}

void get(int u = 1, int p = 0)
{
    dp[u] = 1;
    cnt[u] = 0;
    for(int v : g[u]) if(v - p) {
        get(v, u);

        dp[u] = dp[u] * dp[v] % mod;

        cnt[u] += cnt[v];
        dp[u] *= cmb(cnt[u], cnt[v]);
        dp[u] %= mod;
    }
    cnt[u]++;
}

void dfs(int u = 1, int p = 0, ll upv = 1)
{
    ans[u] = dp[u] * upv % mod;
    ans[u] = ans[u] * cmb(n-1, n-cnt[u]) % mod;
    for(int v : g[u]) if(v - p) {
        ll now = ans[u] * invmod(dp[v]) % mod;
        now = now * invmod(cmb(n-1, n-1-cnt[v])) % mod;
        dfs(v, u, now);
    }
}

int main()
{
    prec();

    cin >> n;
    for(int i=1; i<n; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        g[u].push_back(v);
        g[v].push_back(u);
    }

    get();

    dfs();

    for(int i=1; i<=n; i++) printf("%lld\n", ans[i]);
}
