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 mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define ll long long
#define pb push_back
#define PII pair<int,int>
#define X first
#define Y second
#define MP make_pair
#define PI acos(-1.0)
#define eps 1e-8
const int maxn=3e5+10;
const int mod=1e9+7;
using namespace std;

int n;
ll dp[maxn],sz[maxn],ans[maxn],facn=1;
vector<int>g[maxn];

ll qpow(ll a,ll b)
{
    long long ans=1;
    while(b)
    {
        if(b&1) ans=(ans*a)%mod;
        a=(a*a)%mod;
        b/=2;
    }
    return ans;
}

void dfs1(int x,int pre)
{
    sz[x]=dp[x]=1;
    for(int i=0;i<g[x].size();i++)
    {
        int y=g[x][i];
        if(y!=pre)
        {
            dfs1(y,x);
            sz[x]+=sz[y];
            dp[x]=dp[x]*dp[y]%mod;
        }
    }
    dp[x]=dp[x]*sz[x]%mod;
}

void dfs2(int x,int pre)
{
    ll tmp=1;
    for(int i=0;i<g[x].size();i++)
    {
        int y=g[x][i];
        if(y!=pre)
        {
            tmp=tmp*dp[y]%mod;
        }
    }
    if(x!=1)
        dp[x]=tmp*dp[pre]%mod*(n-sz[x])%mod*qpow(dp[x],mod-2)%mod;
    // printf("%d %d\n",x,dp[x]);
    ans[x]=qpow(dp[x],mod-2)*facn%mod;
    for(int i=0;i<g[x].size();i++)
    {
        int y=g[x][i];
        if(y!=pre)
        {
            // tmp+=dp[y];
            dfs2(y,x);
        }
    }
    
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) facn=facn*i%mod;
    for(int i=0;i<n-1;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        g[x].pb(y);
        g[y].pb(x);
    }
    dfs1(1,0);
    dfs2(1,0);

    for(int i=1;i<=n;i++)
        printf("%lld\n",ans[i]);
    // system("pause");
    return 0;
}