#include<bits/stdc++.h>
using namespace std;

#define int long long

#define pb push_back
#define mp make_pair
#define st first
#define nd second
typedef pair < int , int > pp;
const int mod = 1e9 + 7;
const int N   = 5e3 + 3;

vector < int > V[N];

int sz[N],Cen,T[N],t,M[N],MM[N],C[N][N],fa[N],n,x,y,i,j,k,s,ans;

void f(int x){
    sz[x] = 1;
    int h=1;
    for(int y, i=0; i<V[x].size(); i++)
        if(!sz[ y = V[x][i] ]){
            f(y);
            sz[x] += sz[y];
            if(sz[y] > n/2) h=0;
        }
    if(h && n-sz[x] <= n/2) Cen = x;
}

void g(){
    memset(M , 0 , sizeof M);
    M[0] = 1;
    s = 0;
    for(i=1;i<=t;i++){
        memset(MM , 0 , sizeof MM);
        for(j=0;j<=T[i];j++)
            for(k=0;k<=s;k++)
                MM[j+k] = (MM[j+k] + C[ T[i] ][j] * C[ T[i] ][j] % mod * fa[j] % mod * M[k]) % mod;
    
        s += T[i];
        for(j=0;j<=s;j++) M[j] = MM[j];
    }
    
    for(i=0;i<=s;i++) 
        ans = (ans + (i&1 ? -1 : 1) * fa[s-i] * M[i]) % mod;
}

signed main(){
    cin >> n;
    for(fa[0]=C[0][0]=i=1;i<=n;i++){
        fa[i] = fa[i-1] * i % mod; 
        for(C[i][0]=j=1;j<=i;j++)
            C[i][j] = (C[i-1][j-1] + C[i-1][j]) % mod;
    }
    for(i=2;i<=n;i++){
        cin >> x >> y;
        V[x].pb(y); V[y].pb(x);
    }
    f(1);
    x = Cen;
    for(i=0;i<V[x].size();i++){
        if(sz[x] < sz[ y = V[x][i] ]) T[++t] = n - sz[x];
        else T[++t] = sz[y];
    }
    
    g();
    T[++t] = 1;
    g();
    cout << (ans + mod) % mod;
    return 0;
}
### 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;

#define int long long

#define pb push_back
#define mp make_pair
#define st first
#define nd second
typedef pair < int , int > pp;
const int mod = 1e9 + 7;
const int N   = 5e3 + 3;

vector < int > V[N];

int sz[N],Cen,T[N],t,M[N],MM[N],C[N][N],fa[N],n,x,y,i,j,k,s,ans;

void f(int x){
    sz[x] = 1;
    int h=1;
    for(int y, i=0; i<V[x].size(); i++)
        if(!sz[ y = V[x][i] ]){
            f(y);
            sz[x] += sz[y];
            if(sz[y] > n/2) h=0;
        }
    if(h && n-sz[x] <= n/2) Cen = x;
}

void g(){
    memset(M , 0 , sizeof M);
    M[0] = 1;
    s = 0;
    for(i=1;i<=t;i++){
        memset(MM , 0 , sizeof MM);
        for(j=0;j<=T[i];j++)
            for(k=0;k<=s;k++)
                MM[j+k] = (MM[j+k] + C[ T[i] ][j] * C[ T[i] ][j] % mod * fa[j] % mod * M[k]) % mod;
    
        s += T[i];
        for(j=0;j<=s;j++) M[j] = MM[j];
    }
    
    for(i=0;i<=s;i++) 
        ans = (ans + (i&1 ? -1 : 1) * fa[s-i] * M[i]) % mod;
}

signed main(){
    cin >> n;
    for(fa[0]=C[0][0]=i=1;i<=n;i++){
        fa[i] = fa[i-1] * i % mod; 
        for(C[i][0]=j=1;j<=i;j++)
            C[i][j] = (C[i-1][j-1] + C[i-1][j]) % mod;
    }
    for(i=2;i<=n;i++){
        cin >> x >> y;
        V[x].pb(y); V[y].pb(x);
    }
    f(1);
    x = Cen;
    for(i=0;i<V[x].size();i++){
        if(sz[x] < sz[ y = V[x][i] ]) T[++t] = n - sz[x];
        else T[++t] = sz[y];
    }
    
    g();
    T[++t] = 1;
    g();
    cout << (ans + mod) % mod;
    return 0;
}
```