#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
// typedef __int128 ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define FR first
#define SE second
#define MP make_pair
#define PB push_back
#define vc vector
#define db double
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bin(x) (1ll<<(x))
#define fo(i,l,r) for(int i=(l),I=(r);i<=I;i++)
#define fd(i,r,l) for(int i=(r),I=(l);i>=I;i--)
#define mem(x,val) memset(x,val,sizeof x)
#define Swap(a,b,n) for(int I=0;I<=n;I++) swap(a[I],b[I])
#define PC __builtin_popcountll
#ifdef DEBUG
	#define debug(A,args...) fprintf(stderr,A,##args)
#else
	#define debug(A,args...) printf("")
#endif
#define deb debug("line %d\n",__LINE__)
namespace mine
{
	ll qread()
	{
		ll ans=0,f=1;char c=getchar();
		while(c<'0' or c>'9') {if(c=='-')f=-1;c=getchar();}
		while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar();
		return ans*f;
	}
	void write(ll num){if(num<0) putchar('-'),num=-num;if(num>=10) write(num/10);putchar('0'+num%10);}
	void write1(ll num){write(num);putchar(' ');}
	void write2(ll num){write(num);putchar('\n');}
	template<typename T>inline bool chmax(T&a,const T&b){return a<b?a=b,1:0;}
	template<typename T>inline bool chmin(T&a,const T&b){return a>b?a=b,1:0;}
	ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
	bool IN(ll x,ll l,ll r){return l<=x and x<=r;}
	void GG(){puts("-1");exit(0);}
	
	const db eps=1e-8;
	const int INF=0x3f3f3f3f;
	const int MOD=1e9+7;
	int mm(const int x){return x>=MOD?x-MOD:x;}
	template<typename T> void add(T &x,const int &y){x=(x+y>=MOD?x+y-MOD:x+y);}
	ll qpower(ll x,ll e,int mod=MOD){ll ans=1;while(e){if(e&1)ans=ans*x%mod;x=x*x%mod;e>>=1;}return ans;}
	ll invm(ll x){return qpower(x,MOD-2);}
	const int MM=1e6+10;
	ll fac[MM],facinv[MM],Inv[MM];ll Comb(int n,int m){return n<0 or n<m?0:fac[n]*facinv[m]%MOD*facinv[n-m]%MOD;}
	void PRE()
	{
		fac[0]=1;fo(i,1,MM-1) fac[i]=fac[i-1]*i%MOD;
		facinv[MM-1]=invm(fac[MM-1]);fd(i,MM-1,1) facinv[i-1]=facinv[i]*i%MOD;
		Inv[1]=1;fo(i,2,MM-1) Inv[i]=(MOD-MOD/i)*Inv[MOD%i]%MOD;
	}
	const int N=5e3+10;
	//------------------FIXED------------------
	//求重心
	vc<int> to[N];int siz[N];
	int G,SUM,mxs[N];
	void getrt(int x,int fa)
	{
		mxs[x]=0;siz[x]=1;
		for(auto y:to[x]) if(y!=fa) getrt(y,x),siz[x]+=siz[y],chmax(mxs[x],siz[y]);
		chmax(mxs[x],SUM-siz[x]);if(mxs[x]<mxs[G]) G=x;
	}
	ll f[N],g[N];
	void main()
	{
		int n=qread();fo(i,2,n){ int x=qread(),y=qread();to[x].PB(y),to[y].PB(x); }
		SUM=n,mxs[G=0]=INF;getrt(1,0);debug("G=%d\n",G);
		f[0]=1;int sum=0;ll mul=1;
		for(auto y:to[G])
		{
			int aa=(siz[y]>siz[G]?n-siz[G]:siz[y]);mul=mul*fac[aa]%MOD;
			mem(g,0);
			fo(i,0,sum) fo(j,0,aa) add(g[i+j],f[i]*Comb(aa,j)%MOD*facinv[aa-j]%MOD );
			swap(f,g);sum+=aa;
			deb;
		}ll ans=0;fo(i,0,n-1) add(ans,f[i]*fac[n-i]%MOD*(i&1?MOD-1:1)%MOD );write(ans*mul%MOD);
	}
};//变量重名！
signed main()
{
	#ifdef DEBUG
	//freopen("a.in","r",stdin);
	freopen("z.txt","r",stdin);
	//freopen("a.out","w",stdout);
	#endif
	srand(time(0));
	mine::PRE();
	mine::main();
	debug("\n------------------------------------------\nTime: %.2lf s",1.0*clock()/CLOCKS_PER_SEC);
}
### Prompt

Construct a CPP code solution to the problem outlined:

There is a tree with N vertices. The vertices are numbered 1 through N. The i-th edge (1 \leq i \leq N - 1) connects Vertex x_i and y_i. For vertices v and w (1 \leq v, w \leq N), we will define the distance between v and w d(v, w) as "the number of edges contained in the path v-w".

A squirrel lives in each vertex of the tree. They are planning to move, as follows. First, they will freely choose a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N). Then, for each 1 \leq i \leq N, the squirrel that lived in Vertex i will move to Vertex p_i.

Since they like long travels, they have decided to maximize the total distance they traveled during the process. That is, they will choose p so that d(1, p_1) + d(2, p_2) + ... + d(N, p_N) will be maximized. How many such ways are there to choose p, modulo 10^9 + 7?

Constraints

* 2 \leq N \leq 5,000
* 1 \leq x_i, y_i \leq N
* The input graph is a tree.

Input

Input is given from Standard Input in the following format:


N
x_1 y_1
x_2 y_2
:
x_{N - 1} y_{N - 1}


Output

Print the number of the ways to choose p so that the condition is satisfied, modulo 10^9 + 7.

Examples

Input

3
1 2
2 3


Output

3


Input

4
1 2
1 3
1 4


Output

11


Input

6
1 2
1 3
1 4
2 5
2 6


Output

36


Input

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


Output

396

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
// typedef __int128 ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define FR first
#define SE second
#define MP make_pair
#define PB push_back
#define vc vector
#define db double
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bin(x) (1ll<<(x))
#define fo(i,l,r) for(int i=(l),I=(r);i<=I;i++)
#define fd(i,r,l) for(int i=(r),I=(l);i>=I;i--)
#define mem(x,val) memset(x,val,sizeof x)
#define Swap(a,b,n) for(int I=0;I<=n;I++) swap(a[I],b[I])
#define PC __builtin_popcountll
#ifdef DEBUG
	#define debug(A,args...) fprintf(stderr,A,##args)
#else
	#define debug(A,args...) printf("")
#endif
#define deb debug("line %d\n",__LINE__)
namespace mine
{
	ll qread()
	{
		ll ans=0,f=1;char c=getchar();
		while(c<'0' or c>'9') {if(c=='-')f=-1;c=getchar();}
		while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar();
		return ans*f;
	}
	void write(ll num){if(num<0) putchar('-'),num=-num;if(num>=10) write(num/10);putchar('0'+num%10);}
	void write1(ll num){write(num);putchar(' ');}
	void write2(ll num){write(num);putchar('\n');}
	template<typename T>inline bool chmax(T&a,const T&b){return a<b?a=b,1:0;}
	template<typename T>inline bool chmin(T&a,const T&b){return a>b?a=b,1:0;}
	ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
	bool IN(ll x,ll l,ll r){return l<=x and x<=r;}
	void GG(){puts("-1");exit(0);}
	
	const db eps=1e-8;
	const int INF=0x3f3f3f3f;
	const int MOD=1e9+7;
	int mm(const int x){return x>=MOD?x-MOD:x;}
	template<typename T> void add(T &x,const int &y){x=(x+y>=MOD?x+y-MOD:x+y);}
	ll qpower(ll x,ll e,int mod=MOD){ll ans=1;while(e){if(e&1)ans=ans*x%mod;x=x*x%mod;e>>=1;}return ans;}
	ll invm(ll x){return qpower(x,MOD-2);}
	const int MM=1e6+10;
	ll fac[MM],facinv[MM],Inv[MM];ll Comb(int n,int m){return n<0 or n<m?0:fac[n]*facinv[m]%MOD*facinv[n-m]%MOD;}
	void PRE()
	{
		fac[0]=1;fo(i,1,MM-1) fac[i]=fac[i-1]*i%MOD;
		facinv[MM-1]=invm(fac[MM-1]);fd(i,MM-1,1) facinv[i-1]=facinv[i]*i%MOD;
		Inv[1]=1;fo(i,2,MM-1) Inv[i]=(MOD-MOD/i)*Inv[MOD%i]%MOD;
	}
	const int N=5e3+10;
	//------------------FIXED------------------
	//求重心
	vc<int> to[N];int siz[N];
	int G,SUM,mxs[N];
	void getrt(int x,int fa)
	{
		mxs[x]=0;siz[x]=1;
		for(auto y:to[x]) if(y!=fa) getrt(y,x),siz[x]+=siz[y],chmax(mxs[x],siz[y]);
		chmax(mxs[x],SUM-siz[x]);if(mxs[x]<mxs[G]) G=x;
	}
	ll f[N],g[N];
	void main()
	{
		int n=qread();fo(i,2,n){ int x=qread(),y=qread();to[x].PB(y),to[y].PB(x); }
		SUM=n,mxs[G=0]=INF;getrt(1,0);debug("G=%d\n",G);
		f[0]=1;int sum=0;ll mul=1;
		for(auto y:to[G])
		{
			int aa=(siz[y]>siz[G]?n-siz[G]:siz[y]);mul=mul*fac[aa]%MOD;
			mem(g,0);
			fo(i,0,sum) fo(j,0,aa) add(g[i+j],f[i]*Comb(aa,j)%MOD*facinv[aa-j]%MOD );
			swap(f,g);sum+=aa;
			deb;
		}ll ans=0;fo(i,0,n-1) add(ans,f[i]*fac[n-i]%MOD*(i&1?MOD-1:1)%MOD );write(ans*mul%MOD);
	}
};//变量重名！
signed main()
{
	#ifdef DEBUG
	//freopen("a.in","r",stdin);
	freopen("z.txt","r",stdin);
	//freopen("a.out","w",stdout);
	#endif
	srand(time(0));
	mine::PRE();
	mine::main();
	debug("\n------------------------------------------\nTime: %.2lf s",1.0*clock()/CLOCKS_PER_SEC);
}
```