Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
1 3

Sample Output 1

2
1
1

The graph in this input is as follows:

For k=1, there are two ways in which we can write the numbers on the vertices, as follows:

Writing 1, 2, 3 on Vertex 1, 2, 3, respectively

Writing 1, 3, 2 on Vertex 1, 2, 3, respectively

Sample Input 2

2
1 2

Sample Output 2

1
1

The graph in this input is as follows:

Sample Input 3

5
1 2
2 3
3 4
3 5

Sample Output 3

2
8
12
3
3

The graph in this input is as follows:

Sample Input 4

8
1 2
2 3
3 4
3 5
3 6
6 7
6 8

Sample Output 4

40
280
840
120
120
504
72
72

The graph in this input is as follows:

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define MOD 1000000007
#define INF ((1LL<<62)-(1LL<<31))
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
#define invp(a,p)pom(a,p-2,p)

//階乗,二項係数
//*
ll fact[200010];
ll invfact[200010];
void makefact(int n,ll m){
	fact[0]=1;
	rep(i,0,n)fact[i+1]=fact[i]*(i+1)%m;
	invfact[n]=invp(fact[n],m);
	for(int i=n;i>0;i--)invfact[i-1]=invfact[i]*i%m;
}
ll choose(int n,int r,int m){return n<0||r<0||n-r<0?0:fact[n]*invfact[r]%m*invfact[n-r]%m;}
//*/

//辺の情報を個別に持つタイプ
typedef struct edge{ll s,g,c;}E;
typedef struct graph{
	int vcnt,ecnt;
	E  e[400010];//適宜変える(ecnt)
	int inv[400010];//逆辺のindex(ecnt)
	int id[200010];//適宜変える(vcnt)
}G;
G g;

int esort(const void*a,const void*b){
	E*p=(E*)a,*q=(E*)b;
	if((*p).s<(*q).s)return -1;
	if((*p).s>(*q).s)return  1;
	if((*p).g<(*q).g)return -1;
	return 1;
}
void makeinvedge();
void readgraph(){
	//適宜変える
	ll n;
	scanf("%lld",&n);
	rep(i,0,n-1){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		x--,y--;
		g.e[i].s=x;
		g.e[i].g=y;
	}
	g.vcnt=n;
	g.ecnt=n-1;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//番兵
}

void makeinvedge(){
	//逆辺とidx
	int added=0;
	rep(i,0,g.ecnt){
		int l=0,r=g.ecnt;
		while(r-l>1){
			int m=(l+r)/2;
			if(g.e[m].s<g.e[i].g||(g.e[m].s==g.e[i].g&&g.e[m].g<=g.e[i].s))l=m;
			else r=m;
		}
		if(g.e[l].s!=g.e[i].g||g.e[l].g!=g.e[i].s){
			g.e[g.ecnt+added].s=g.e[i].g;
			g.e[g.ecnt+added].g=g.e[i].s;
			g.e[g.ecnt+added].c=0;
			added++;
		}
	}
	g.ecnt+=added;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//番兵

	rep(i,0,g.ecnt){
		int l=0,r=g.ecnt;
		while(r-l>1){
			int m=(l+r)/2;
			if(g.e[m].s<g.e[i].g||(g.e[m].s==g.e[i].g&&g.e[m].g<=g.e[i].s))l=m;
			else r=m;
		}
		g.inv[i]=l;
	}
}

//†全方位木DP†
//*
typedef struct rerooting_atai{ll n,c;}atai;
atai dpv[200010];//頂点に対する答え
atai dpe[400010];//辺に対する答え
atai cum[2][400010];//累積和
atai munit={0,1};//単位元
atai init={-1,0};
int isinit(atai x){return x.n==init.n;}
atai fold_e(atai x,atai y){
	atai ret;
	ret.n=x.n+y.n;
	ret.c=x.c*y.c%MOD*choose(x.n+y.n,x.n,MOD)%MOD;
	return ret;
}
atai add_v(atai x,int v){x.n++;return x;}
atai add_e(atai x,int e){return x;}

void rrinit(){
	makeinvedge();
	rep(i,0,g.vcnt)dpv[i]=init;
	rep(i,0,g.ecnt)dpe[i]=init;
}
atai dfs_e(int);
atai dfs_v(int v){
	if(!isinit(dpv[v]))return dpv[v];
	rep(i,g.id[v],g.id[v+1]){
		if(i==g.id[v])cum[0][i]=munit;
		else cum[0][i]=fold_e(cum[0][i-1],dfs_e(i-1));
	}
	for(int i=g.id[v+1]-1;i>=g.id[v];i--){
		if(i==g.id[v+1]-1)cum[1][i]=munit;
		else cum[1][i]=fold_e(cum[1][i+1],dfs_e(i+1));
	}
	atai ret=munit;
	rep(i,g.id[v],g.id[v+1])ret=fold_e(ret,dfs_e(i));
	ret=add_v(ret,v);
	return dpv[v]=ret;
}
atai dfs_e(int e){
	if(!isinit(dpe[e]))return dpe[e];
	atai ret;
	if(!isinit(dpv[g.e[e].g])){
		ret=fold_e(cum[0][g.inv[e]],cum[1][g.inv[e]]);
	}else{
		ret=munit;
		int v=g.e[e].g;
		rep(i,g.id[v],g.id[v+1])if(g.e[i].g!=g.e[e].s)ret=fold_e(ret,dfs_e(i));
	}
	ret=add_v(ret,g.e[e].g);
	ret=add_e(ret,e);
	return dpe[e]=ret;
}
void rerootingDP(int v,int pre){
	if(pre==-1)rrinit();
	dfs_v(v);
	rep(i,g.id[v],g.id[v+1])if(g.e[i].g!=pre)rerootingDP(g.e[i].g,v);
}
//*/

int main(){
	readgraph();
	makefact(g.vcnt+5,MOD);
	rerootingDP(0,-1);
	rep(i,0,g.vcnt)printf("%lld\n",dpv[i].c);
}