#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <algorithm>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define repi(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define repdi(i,a,b) for(ll i=(a)-1;i>=(b);--i)
#define repd(i,a) repdi(i,a,0)
#define itr(it,a) for( auto it = (a).begin(); it != (a).end(); ++it )
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()

using ll = long long;
using P = std::pair<ll, ll>;

constexpr ll INF = 1ll<<60;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class S, class T>
std::ostream& operator<< ( std::ostream& out, const std::pair<S,T>& a )
{ std::cout << '(' << a.first << ", " << a.second << ')'; return out; }

template<class T>
std::ostream &operator<< ( std::ostream& out, const std::vector<T>& a )
{ std::cout << '['; rep( i, a.size() ){ std::cout << a[i]; if( i != a.size()-1 ) std::cout << ", "; } std::cout << ']'; return out; }

const ll mod = 1000000007;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

struct combination {
  std::vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
};

ll N;
std::vector<ll> G[5010];
std::vector<ll> centroids;
ll subtree[5010], par[5010];
combination comb( 10010 );

void dfs( ll v, ll p = -1 )
{
  bool isCentroid = true;

  subtree[v] = 1;

  for( auto u : G[v] ) if( u != p )
  {
    dfs( u, v );
    par[u] = v;
    subtree[v] += subtree[u];

    if( 2*subtree[u] > N )
      isCentroid = false;
  }

  if( 2*(N-subtree[v]) > N )
    isCentroid = false;

  if( isCentroid )
    centroids.emplace_back( v );

  return;
}

std::vector<ll> ts;
mint dp[5010][5010];

int main()
{
  std::cin >> N;

  rep( i, N-1 )
  {
    ll x, y;
    std::cin >> x >> y;
    --x; --y;

    G[x].emplace_back( y );
    G[y].emplace_back( x );
  }

  dfs( 0 );

  if( centroids.size() == 2 )
  {
    std::cout << comb.fact[N/2].pow(2).x << std::endl;

    return 0;
  }

  ll c = centroids[0];

  ts.emplace_back( N-subtree[c] );

  for( auto u : G[c] ) if( par[c] != u )
    ts.emplace_back( subtree[u] );

  dp[0][0] = 1;

  rep( i, ts.size() ) rep( j, N+1 )
  {
    rep( x, ts[i]+1 )
      dp[i+1][j+x] += dp[i][j]*comb( ts[i], x ).pow(2)*comb.fact[x];
  }

  mint ans = 0;

  rep( j, N+1 )
    ans += dp[ts.size()][j]*comb.fact[N-j]*(j&1 ? -1 : 1);

  std::cout << ans.x << std::endl;

  return 0;
}
### Prompt

Generate 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
#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <algorithm>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define repi(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define repdi(i,a,b) for(ll i=(a)-1;i>=(b);--i)
#define repd(i,a) repdi(i,a,0)
#define itr(it,a) for( auto it = (a).begin(); it != (a).end(); ++it )
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()

using ll = long long;
using P = std::pair<ll, ll>;

constexpr ll INF = 1ll<<60;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class S, class T>
std::ostream& operator<< ( std::ostream& out, const std::pair<S,T>& a )
{ std::cout << '(' << a.first << ", " << a.second << ')'; return out; }

template<class T>
std::ostream &operator<< ( std::ostream& out, const std::vector<T>& a )
{ std::cout << '['; rep( i, a.size() ){ std::cout << a[i]; if( i != a.size()-1 ) std::cout << ", "; } std::cout << ']'; return out; }

const ll mod = 1000000007;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

struct combination {
  std::vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
};

ll N;
std::vector<ll> G[5010];
std::vector<ll> centroids;
ll subtree[5010], par[5010];
combination comb( 10010 );

void dfs( ll v, ll p = -1 )
{
  bool isCentroid = true;

  subtree[v] = 1;

  for( auto u : G[v] ) if( u != p )
  {
    dfs( u, v );
    par[u] = v;
    subtree[v] += subtree[u];

    if( 2*subtree[u] > N )
      isCentroid = false;
  }

  if( 2*(N-subtree[v]) > N )
    isCentroid = false;

  if( isCentroid )
    centroids.emplace_back( v );

  return;
}

std::vector<ll> ts;
mint dp[5010][5010];

int main()
{
  std::cin >> N;

  rep( i, N-1 )
  {
    ll x, y;
    std::cin >> x >> y;
    --x; --y;

    G[x].emplace_back( y );
    G[y].emplace_back( x );
  }

  dfs( 0 );

  if( centroids.size() == 2 )
  {
    std::cout << comb.fact[N/2].pow(2).x << std::endl;

    return 0;
  }

  ll c = centroids[0];

  ts.emplace_back( N-subtree[c] );

  for( auto u : G[c] ) if( par[c] != u )
    ts.emplace_back( subtree[u] );

  dp[0][0] = 1;

  rep( i, ts.size() ) rep( j, N+1 )
  {
    rep( x, ts[i]+1 )
      dp[i+1][j+x] += dp[i][j]*comb( ts[i], x ).pow(2)*comb.fact[x];
  }

  mint ans = 0;

  rep( j, N+1 )
    ans += dp[ts.size()][j]*comb.fact[N-j]*(j&1 ? -1 : 1);

  std::cout << ans.x << std::endl;

  return 0;
}
```