#include<bits/stdc++.h>
using namespace std;
int n;
int u,v;
vector<int> ga[200100];
vector<int> ch[200100];
int zsize[200100];
long long dp1[200100],dp2[200100],MOD=1000000007;
long long fac[200100],ifac[200100],inv[200100];
long long C(long long x,long long y){
	return fac[x]*ifac[y]%MOD*ifac[x-y]%MOD;
}
long long kuai(long long x,long long y){
	long long rr=1LL;
	while(y){
		if(y&1)
			rr=(rr*x)%MOD;
		x=(x*x)%MOD;
		y>>=1;
	}
	return rr;
}
void solve1(int u,int p){
	dp1[u]=1LL;
	for(int v:ga[u]){
		if(v!=p){
			ch[u].push_back(v);
			solve1(v,u);
			zsize[u]+=zsize[v];
			dp1[u]=dp1[u]*C(zsize[u],zsize[v])%MOD*dp1[v]%MOD;
		}
	}
	zsize[u]++;
}
void solve2(int u){
	for(int v:ch[u]){
		dp2[v]=dp2[u]*dp1[v]%MOD*kuai(dp1[v]*C(n-1,zsize[v])%MOD,MOD-2)%MOD*C(n-1,n-zsize[v])%MOD;
		solve2(v);
	}
}
int main(){
	#ifdef FILIN
		#ifndef LOCAL
			freopen(FILIN,"r",stdin);
			freopen(FILOUT,"w",stdout);
		#endif
	#endif
	cin>>n;
	inv[1]=1LL;
	for(int i=2;i<=n;i++)
		inv[i]=(MOD-((MOD/i)*inv[MOD%i])%MOD);
	fac[0]=1LL;
	ifac[0]=1LL; 
	for(int i=1;i<=n;i++){
		fac[i]=(fac[i-1]*(long long)(i))%MOD;
		ifac[i]=(ifac[i-1]*inv[i])%MOD;
	}
	for(int i=1;i<n;i++){
		cin>>u>>v;
		ga[u].push_back(v);
		ga[v].push_back(u);
	} 
	solve1(1,-1);
	dp2[1]=dp1[1];
	solve2(1);
	for(int i=1;i<=n;i++){
		cout<<dp2[i]<<endl;
	}
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int n;
int u,v;
vector<int> ga[200100];
vector<int> ch[200100];
int zsize[200100];
long long dp1[200100],dp2[200100],MOD=1000000007;
long long fac[200100],ifac[200100],inv[200100];
long long C(long long x,long long y){
	return fac[x]*ifac[y]%MOD*ifac[x-y]%MOD;
}
long long kuai(long long x,long long y){
	long long rr=1LL;
	while(y){
		if(y&1)
			rr=(rr*x)%MOD;
		x=(x*x)%MOD;
		y>>=1;
	}
	return rr;
}
void solve1(int u,int p){
	dp1[u]=1LL;
	for(int v:ga[u]){
		if(v!=p){
			ch[u].push_back(v);
			solve1(v,u);
			zsize[u]+=zsize[v];
			dp1[u]=dp1[u]*C(zsize[u],zsize[v])%MOD*dp1[v]%MOD;
		}
	}
	zsize[u]++;
}
void solve2(int u){
	for(int v:ch[u]){
		dp2[v]=dp2[u]*dp1[v]%MOD*kuai(dp1[v]*C(n-1,zsize[v])%MOD,MOD-2)%MOD*C(n-1,n-zsize[v])%MOD;
		solve2(v);
	}
}
int main(){
	#ifdef FILIN
		#ifndef LOCAL
			freopen(FILIN,"r",stdin);
			freopen(FILOUT,"w",stdout);
		#endif
	#endif
	cin>>n;
	inv[1]=1LL;
	for(int i=2;i<=n;i++)
		inv[i]=(MOD-((MOD/i)*inv[MOD%i])%MOD);
	fac[0]=1LL;
	ifac[0]=1LL; 
	for(int i=1;i<=n;i++){
		fac[i]=(fac[i-1]*(long long)(i))%MOD;
		ifac[i]=(ifac[i-1]*inv[i])%MOD;
	}
	for(int i=1;i<n;i++){
		cin>>u>>v;
		ga[u].push_back(v);
		ga[v].push_back(u);
	} 
	solve1(1,-1);
	dp2[1]=dp1[1];
	solve2(1);
	for(int i=1;i<=n;i++){
		cout<<dp2[i]<<endl;
	}
	return 0;
}

```