// This amazing code is by Eric Sunli Chen.
#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <utility>
#include <vector>
using namespace std;
template<typename T> bool get_int(T &x)
{
	char t=getchar();
	bool neg=false;
	x=0;
	for(; (t>'9'||t<'0')&&t!='-'&&t!=EOF; t=getchar());
	if(t=='-')neg=true,t=getchar();if(t==EOF)return false;
	for(; t<='9'&&t>='0'; t=getchar())x=x*10+t-'0';
	if(neg)x=-x;return true;
}
template<typename T> void print_int(T x)
{
	if(x<0)putchar('-'),x=-x;
	short a[20]= {},sz=0;
	while(x>0)a[sz++]=x%10,x/=10;
	if(sz==0)putchar('0');
	for(int i=sz-1; i>=0; i--)putchar('0'+a[i]);
}
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define get1(a) get_int(a)
#define get2(a,b) (get1(a)&&get1(b))
#define get3(a,b,c) (get1(a)&&get2(b,c))
#define printendl(a) print_int(a),puts("")
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const LL Linf=1ll<<61;
const double pi=acos(-1.0);

const int mod=1e9+7;

int power(int x,int y){int ret=1;do{if(y&1)ret=1ll*ret*x%mod;x=1ll*x*x%mod;}while(y>>=1);return ret;}
int f[5111],invf[5111];

vector<int>g[5111];
int n,fa[5111],sz[5111],mx[5111];
void dfs(int x,int f=0)
{
	fa[x]=f;sz[x]=1;
	for(auto v:g[x])if(v!=f){dfs(v,x);sz[x]+=sz[v];mx[x]=max(mx[x],sz[v]);}
	mx[x]=max(mx[x],n-sz[x]);
}

int dp[5111],tsz;
void add(int v)
{
	for(int i=tsz;i>=0;i--)
	{
		for(int j=1;j<=v;j++)
			dp[i+j]=(dp[i+j]+1ll*dp[i]*f[v]%mod*invf[v-j]%mod*f[v]%mod*invf[j]%mod*invf[v-j])%mod;
	}
	tsz+=v;
}
int main()
{
	f[0]=invf[0]=1;for(int i=1;i<5111;i++){f[i]=1ll*i*f[i-1]%mod;invf[i]=power(f[i],mod-2);}
	get1(n);
	for(int i=1,u,v;i<n;i++)
	{
		get2(u,v);
		g[u].pb(v);g[v].pb(u);
	}
	dfs(1);
	int rt=min_element(mx+1,mx+n+1)-mx;
	for(int i=1;i<=n;i++)if(mx[rt]==mx[i]&&rt!=i)
	{
		printf("%d\n",1ll*f[n/2]*f[n/2]%mod);
		return 0;
	}
	dp[0]=1;add(n-sz[rt]);
	for(auto v:g[rt])if(v!=fa[rt])add(sz[v]);
	int ans=0;
	for(int i=0;i<n;i++)
	{
		if(i&1)ans=(ans-1ll*dp[i]*f[n-i]%mod+mod)%mod;
		else ans=(ans+1ll*dp[i]*f[n-i])%mod;
	}
	printendl(ans);
	return 0;
}
### Prompt

Your challenge is to write a cpp solution to the following problem:

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
// This amazing code is by Eric Sunli Chen.
#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <utility>
#include <vector>
using namespace std;
template<typename T> bool get_int(T &x)
{
	char t=getchar();
	bool neg=false;
	x=0;
	for(; (t>'9'||t<'0')&&t!='-'&&t!=EOF; t=getchar());
	if(t=='-')neg=true,t=getchar();if(t==EOF)return false;
	for(; t<='9'&&t>='0'; t=getchar())x=x*10+t-'0';
	if(neg)x=-x;return true;
}
template<typename T> void print_int(T x)
{
	if(x<0)putchar('-'),x=-x;
	short a[20]= {},sz=0;
	while(x>0)a[sz++]=x%10,x/=10;
	if(sz==0)putchar('0');
	for(int i=sz-1; i>=0; i--)putchar('0'+a[i]);
}
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define get1(a) get_int(a)
#define get2(a,b) (get1(a)&&get1(b))
#define get3(a,b,c) (get1(a)&&get2(b,c))
#define printendl(a) print_int(a),puts("")
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const LL Linf=1ll<<61;
const double pi=acos(-1.0);

const int mod=1e9+7;

int power(int x,int y){int ret=1;do{if(y&1)ret=1ll*ret*x%mod;x=1ll*x*x%mod;}while(y>>=1);return ret;}
int f[5111],invf[5111];

vector<int>g[5111];
int n,fa[5111],sz[5111],mx[5111];
void dfs(int x,int f=0)
{
	fa[x]=f;sz[x]=1;
	for(auto v:g[x])if(v!=f){dfs(v,x);sz[x]+=sz[v];mx[x]=max(mx[x],sz[v]);}
	mx[x]=max(mx[x],n-sz[x]);
}

int dp[5111],tsz;
void add(int v)
{
	for(int i=tsz;i>=0;i--)
	{
		for(int j=1;j<=v;j++)
			dp[i+j]=(dp[i+j]+1ll*dp[i]*f[v]%mod*invf[v-j]%mod*f[v]%mod*invf[j]%mod*invf[v-j])%mod;
	}
	tsz+=v;
}
int main()
{
	f[0]=invf[0]=1;for(int i=1;i<5111;i++){f[i]=1ll*i*f[i-1]%mod;invf[i]=power(f[i],mod-2);}
	get1(n);
	for(int i=1,u,v;i<n;i++)
	{
		get2(u,v);
		g[u].pb(v);g[v].pb(u);
	}
	dfs(1);
	int rt=min_element(mx+1,mx+n+1)-mx;
	for(int i=1;i<=n;i++)if(mx[rt]==mx[i]&&rt!=i)
	{
		printf("%d\n",1ll*f[n/2]*f[n/2]%mod);
		return 0;
	}
	dp[0]=1;add(n-sz[rt]);
	for(auto v:g[rt])if(v!=fa[rt])add(sz[v]);
	int ans=0;
	for(int i=0;i<n;i++)
	{
		if(i&1)ans=(ans-1ll*dp[i]*f[n-i]%mod+mod)%mod;
		else ans=(ans+1ll*dp[i]*f[n-i])%mod;
	}
	printendl(ans);
	return 0;
}
```