#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=5005,mod=1e9+7;
ll f[N],invf[N];
ll C(int n,int m)
{
    return f[n]*invf[n-m]%mod*invf[m]%mod;
}
int n,tot,rt,si[N],g[N],head[N],nex[N<<1],to[N<<1];
bool two;
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
void getrt(int u,int p)
{
    si[u]=1;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        getrt(v,u);
        si[u]+=si[v];
        g[u]=max(g[u],si[v]);
    }
    g[u]=max(g[u],n-si[u]);
    if(g[u]<g[rt]) rt=u,two=false;
    else if(g[u]==g[rt]) two=true;
}
void dfs(int u,int p)
{
    si[u]=1;
    for(int i=head[u];i;i=nex[i])
        if(to[i]!=p) dfs(to[i],u),si[u]+=si[to[i]];
}
int top,s[N];
ll dp[N][N];
int main()
{
    f[0]=invf[0]=f[1]=invf[1]=1;
    for(int i=2;i<N;i++) f[i]=f[i-1]*i%mod,invf[i]=(mod-mod/i)*invf[mod%i]%mod;
    for(int i=2;i<N;i++) invf[i]=invf[i]*invf[i-1]%mod;
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int u,v;scanf("%d%d",&u,&v);
        add(u,v);add(v,u);
    }
    g[rt]=1e9;
    getrt(1,0);
    if(two)
    {
        assert(n%2==0);
        printf("%lld\n",f[n/2]*f[n/2]%mod);
        return 0;
    }
    for(int i=head[rt];i;i=nex[i])
    {
        int v=to[i];dfs(v,rt);
        s[++top]=si[v];
    }
    sort(s+1,s+1+top);
    dp[0][0]=1;
    int sum=0;
    for(int i=0;i<top;i++)
    {
        for(int j=0;j<=sum;j++)
            for(int k=0;k<=s[i+1];k++)
                (dp[i+1][j+k]+=dp[i][j]*C(s[i+1],k)%mod*C(s[i+1],k)%mod*f[k]%mod)%=mod;
        sum+=s[i+1];
    }
    sum++;
    ll ans=0;
    for(int s=1,i=0;i<=sum;i++,s*=-1)
        (ans+=s*dp[top][i]*f[sum-i])%=mod;
    ans=(ans+mod)%mod;
    printf("%lld\n",ans);
}

### Prompt

In cpp, your task is to solve 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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=5005,mod=1e9+7;
ll f[N],invf[N];
ll C(int n,int m)
{
    return f[n]*invf[n-m]%mod*invf[m]%mod;
}
int n,tot,rt,si[N],g[N],head[N],nex[N<<1],to[N<<1];
bool two;
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
void getrt(int u,int p)
{
    si[u]=1;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        getrt(v,u);
        si[u]+=si[v];
        g[u]=max(g[u],si[v]);
    }
    g[u]=max(g[u],n-si[u]);
    if(g[u]<g[rt]) rt=u,two=false;
    else if(g[u]==g[rt]) two=true;
}
void dfs(int u,int p)
{
    si[u]=1;
    for(int i=head[u];i;i=nex[i])
        if(to[i]!=p) dfs(to[i],u),si[u]+=si[to[i]];
}
int top,s[N];
ll dp[N][N];
int main()
{
    f[0]=invf[0]=f[1]=invf[1]=1;
    for(int i=2;i<N;i++) f[i]=f[i-1]*i%mod,invf[i]=(mod-mod/i)*invf[mod%i]%mod;
    for(int i=2;i<N;i++) invf[i]=invf[i]*invf[i-1]%mod;
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int u,v;scanf("%d%d",&u,&v);
        add(u,v);add(v,u);
    }
    g[rt]=1e9;
    getrt(1,0);
    if(two)
    {
        assert(n%2==0);
        printf("%lld\n",f[n/2]*f[n/2]%mod);
        return 0;
    }
    for(int i=head[rt];i;i=nex[i])
    {
        int v=to[i];dfs(v,rt);
        s[++top]=si[v];
    }
    sort(s+1,s+1+top);
    dp[0][0]=1;
    int sum=0;
    for(int i=0;i<top;i++)
    {
        for(int j=0;j<=sum;j++)
            for(int k=0;k<=s[i+1];k++)
                (dp[i+1][j+k]+=dp[i][j]*C(s[i+1],k)%mod*C(s[i+1],k)%mod*f[k]%mod)%=mod;
        sum+=s[i+1];
    }
    sum++;
    ll ans=0;
    for(int s=1,i=0;i<=sum;i++,s*=-1)
        (ans+=s*dp[top][i]*f[sum-i])%=mod;
    ans=(ans+mod)%mod;
    printf("%lld\n",ans);
}

```