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>
#define ll long long
#define MAX 500005
#define P ((ll)1e9+7)
using namespace std;

ll qpow(ll a, ll n){
    ll res = 1;
    while(n){
        if(n&1) res = res*a%P;
        a = a*a%P, n >>= 1;
    }
    return res;
}
ll inv(ll x){
    return qpow(x, P-2);
}

int n, cnt;
int head[MAX], vet[MAX], Next[MAX];
ll f[MAX], sz[MAX], fac[MAX], g[MAX];

void init(){
    fac[0] = 1;
    for(int i = 1; i < MAX; i++){
        fac[i] = fac[i-1]*i%P;
    }
}

void add(int x, int y){
    cnt++;
    Next[cnt] = head[x];
    head[x] = cnt;
    vet[cnt] = y;
}

void dfs(int x, int fa){
    sz[x] = 1;
    for(int i = head[x]; i; i = Next[i]){
        int v = vet[i];
        if(v == fa) continue;
        dfs(v, x);
        sz[x] += sz[v];
    }
    f[x] = fac[sz[x]-1];
    for(int i = head[x]; i; i = Next[i]){
        int v = vet[i];
        if(v == fa) continue;
        f[x] = f[x]*inv(fac[sz[v]])%P*f[v]%P;
    }
    g[x] = f[x];
}

void dfs2(int x, int fa){
    for(int i = head[x]; i; i = Next[i]){
        int v = vet[i];
        if(v == fa) continue;
        ll t = fac[n-1]*inv(fac[n-1-sz[v]])%P, d = fac[n-1]*inv(fac[sz[v]-1])%P;
        ll F = g[x]*inv(t)%P*fac[sz[v]]%P*inv(f[v])%P;
        g[v] = g[v]*F%P*d%P*inv(fac[n-sz[v]])%P;
        dfs2(v, x);
    }
}

int main()
{
    init();
    cin >> n;
    int x, y;
    for(int i = 1; i < n; i++){
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }
    dfs(1, 0);
    g[1] = f[1];
    dfs2(1, 0);
    for(int i = 1; i <= n; i++){
        printf("%lld\n", g[i]);
    }

    return 0;
}
