#include <bits/stdc++.h>
#define ADD(a, b) a = (a + ll(b)) % mod
#define MUL(a, b) a = (a * ll(b)) % mod
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)
#define all(a) (a).begin(), (a).end()
#define sz(v) (int)(v).size()
#define pb push_back
#define sec second
#define fst first
#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<int, pi> ppi;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> mat;
typedef complex<double> comp;
void Debug() {cout << '\n'; }
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
	cout<<arg<<" ";Debug(rest...);}
template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {
	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}
template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){
	out<<"("<<v.first<<", "<<v.second<<")";return out;}
const int MAX_N = 5010;
const int MAX_V = 100010;
const double eps = 1e-6;
const ll mod = 1000000007;
const int inf = 1 << 29;
const ll linf = 1LL << 60;
const double PI = 3.14159265358979323846;
///////////////////////////////////////////////////////////////////////////////////////////////////

ll fac[MAX_N], inv[MAX_N], fiv[MAX_N]; //fiv:inv(fac(i))

ll mod_pow(ll a, ll n) {
	if(n == 0) return 1;
	ll res = mod_pow(a, n / 2);
	if(n % 2 == 0) return res * res % mod;
	else return a * res % mod * res % mod;
}

ll invf(ll a) {
	return mod_pow(a, mod - 2);
}

void C_init(int n) {
	fac[0] = fac[1] = 1; inv[1] = 1;
	fiv[0] = fiv[1] = 1;
	rep(i, 2, n + 1) {
		fac[i] = fac[i - 1] * i % mod;
		inv[i] = mod - inv[mod % i] * (mod / i) % mod;
		fiv[i] = fiv[i - 1] * inv[i] % mod;
	}
}

ll C(int a, int b) { //assume a >= b
	if(a < b || a < 0 || b < 0) return 0;
	return fac[a] * fiv[b] % mod * fiv[a - b] % mod;
}

int N;
vector<int> cen;
vector<int> G[MAX_N];
ll dp[2][MAX_N];

int search_centroid(int v, int p) {
	int s = 1, m = 0;
	rep(i, 0, sz(G[v])) {
		int n = G[v][i];
		if(p == n) continue;
		int t = search_centroid(n, v);
		s += t;
		MAX(m, t);
	}
	MAX(m, N - s);
	if(m <= N / 2) {
		cen.pb(v);
	}
	return s;
}

void solve() {
	cin >> N;
	C_init(N);
	rep(i, 0, N - 1) {
		int a, b; cin >> a >> b; a--; b--;
		G[b].pb(a);
		G[a].pb(b);
	}
	search_centroid(0, -1);
	if(sz(cen) == 2) {
		cout << fac[N / 2] * fac[N / 2] % mod << "\n";
	}
	else if(sz(cen) == 1) {
		int c = cen[0];
		vector<int> vec;
		rep(i, 0, sz(G[c])) {
			int n = G[c][i];
			vec.pb(search_centroid(n, c));
		}
		int K = sz(vec);
		dp[0][0] = 1;
		int s = 0;
		int now = 0, next = 1;
		rep(i, 0, K) {
			memset(dp[next], 0, sizeof(dp[next]));
			rep(j, 0, s + 1) {
				if(!dp[now][j]) continue;
				int t = vec[i];
				rep(k, 0, t + 1) {
					ADD(dp[next][j + k], dp[now][j] * C(t, k) % mod * C(t, k) % mod * fac[k] % mod);
				}
			}
			s += vec[i];
			swap(now, next);
		}
		ll ans = 0;
		rep(i, 0, N) {
			if(i % 2) ADD(ans, mod - fac[N - i] * dp[now][i] % mod);
			else ADD(ans, fac[N - i] * dp[now][i] % mod);
		}
		cout << ans << "\n";
	}
}

int main() {
#ifndef LOCAL
	ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    cout << fixed;
	cout.precision(20);
#ifdef LOCAL
    freopen("in.txt", "rt", stdin);
#endif	
	solve();
#ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
	return 0;
}


### Prompt

Create a solution in cpp for 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>
#define ADD(a, b) a = (a + ll(b)) % mod
#define MUL(a, b) a = (a * ll(b)) % mod
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)
#define all(a) (a).begin(), (a).end()
#define sz(v) (int)(v).size()
#define pb push_back
#define sec second
#define fst first
#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<int, pi> ppi;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> mat;
typedef complex<double> comp;
void Debug() {cout << '\n'; }
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
	cout<<arg<<" ";Debug(rest...);}
template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {
	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}
template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){
	out<<"("<<v.first<<", "<<v.second<<")";return out;}
const int MAX_N = 5010;
const int MAX_V = 100010;
const double eps = 1e-6;
const ll mod = 1000000007;
const int inf = 1 << 29;
const ll linf = 1LL << 60;
const double PI = 3.14159265358979323846;
///////////////////////////////////////////////////////////////////////////////////////////////////

ll fac[MAX_N], inv[MAX_N], fiv[MAX_N]; //fiv:inv(fac(i))

ll mod_pow(ll a, ll n) {
	if(n == 0) return 1;
	ll res = mod_pow(a, n / 2);
	if(n % 2 == 0) return res * res % mod;
	else return a * res % mod * res % mod;
}

ll invf(ll a) {
	return mod_pow(a, mod - 2);
}

void C_init(int n) {
	fac[0] = fac[1] = 1; inv[1] = 1;
	fiv[0] = fiv[1] = 1;
	rep(i, 2, n + 1) {
		fac[i] = fac[i - 1] * i % mod;
		inv[i] = mod - inv[mod % i] * (mod / i) % mod;
		fiv[i] = fiv[i - 1] * inv[i] % mod;
	}
}

ll C(int a, int b) { //assume a >= b
	if(a < b || a < 0 || b < 0) return 0;
	return fac[a] * fiv[b] % mod * fiv[a - b] % mod;
}

int N;
vector<int> cen;
vector<int> G[MAX_N];
ll dp[2][MAX_N];

int search_centroid(int v, int p) {
	int s = 1, m = 0;
	rep(i, 0, sz(G[v])) {
		int n = G[v][i];
		if(p == n) continue;
		int t = search_centroid(n, v);
		s += t;
		MAX(m, t);
	}
	MAX(m, N - s);
	if(m <= N / 2) {
		cen.pb(v);
	}
	return s;
}

void solve() {
	cin >> N;
	C_init(N);
	rep(i, 0, N - 1) {
		int a, b; cin >> a >> b; a--; b--;
		G[b].pb(a);
		G[a].pb(b);
	}
	search_centroid(0, -1);
	if(sz(cen) == 2) {
		cout << fac[N / 2] * fac[N / 2] % mod << "\n";
	}
	else if(sz(cen) == 1) {
		int c = cen[0];
		vector<int> vec;
		rep(i, 0, sz(G[c])) {
			int n = G[c][i];
			vec.pb(search_centroid(n, c));
		}
		int K = sz(vec);
		dp[0][0] = 1;
		int s = 0;
		int now = 0, next = 1;
		rep(i, 0, K) {
			memset(dp[next], 0, sizeof(dp[next]));
			rep(j, 0, s + 1) {
				if(!dp[now][j]) continue;
				int t = vec[i];
				rep(k, 0, t + 1) {
					ADD(dp[next][j + k], dp[now][j] * C(t, k) % mod * C(t, k) % mod * fac[k] % mod);
				}
			}
			s += vec[i];
			swap(now, next);
		}
		ll ans = 0;
		rep(i, 0, N) {
			if(i % 2) ADD(ans, mod - fac[N - i] * dp[now][i] % mod);
			else ADD(ans, fac[N - i] * dp[now][i] % mod);
		}
		cout << ans << "\n";
	}
}

int main() {
#ifndef LOCAL
	ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    cout << fixed;
	cout.precision(20);
#ifdef LOCAL
    freopen("in.txt", "rt", stdin);
#endif	
	solve();
#ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
	return 0;
}


```