#include <cstdio>
#include <algorithm>
#include <vector>
#include <functional>
#include <cassert>
#include <map>
#include <set>
using namespace std;

const int MOD = 1e9 + 7;

const int N = 5000001;
long long fact[N];
long long invfact[N];
long long inv[N];
void init() {
        fact[0] = fact[1] = 1;
        for (int i = 2; i < N; i ++) fact[i] = fact[i - 1] * i % MOD;
        inv[1] = 1;
        for (int i = 2; i < N; i ++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
        invfact[0] = invfact[1] = 1;
        for (int i = 2; i < N; i ++) invfact[i] = invfact[i - 1] * inv[i] % MOD;
}
long long C(long long n, long long r) {
        if (n < 0 || r < 0 || n < r) return 0;
        return fact[n] * invfact[n - r] % MOD * invfact[r] % MOD;
}

vector<int> get_centroids(vector<vector<int> > const & g, int root, set<int> const & forbidden) {
    map<int, int> available; {
        function<void (int, int)> go = [&](int i, int parent) {
            available.emplace(i, available.size());
            for (auto j : g[i]) if (j != parent and not forbidden.count(j)) {
                go(j, i);
            }
        };
        go(root, -1);
    }
    int n = available.size();
    vector<int> result;
    vector<int> size(n, -1);
    function<void (int, int)> go = [&](int x, int parent) {
        bool is_centroid = true;
        int i = available[x];
        size[i] = 1;
        for (auto y : g[x]) if (y != parent and !forbidden.count(y)) {
            int j = available[y];
            go(y, x);
            size[i] += size[j];
            if (size[j] > n / 2) is_centroid = false;
        }
        if (n - size[i] > n / 2) is_centroid = false;
        if (is_centroid) result.push_back(x);
    };
    go(root, -1);
    return result;
}

int main() {
        init();
        int n;
        scanf("%d", &n);
        vector<vector<int>> g(n);
        for (int i = 0; i < n - 1; i ++) {
                int a, b;
                scanf("%d%d", &a, &b);
                a --, b --;
                g[a].push_back(b);
                g[b].push_back(a);
        }
        set<int> forbidden;
        auto get = get_centroids(g, 0, forbidden);
        if (get.size() == 2) {
                long long ans = fact[n / 2] * fact[n / 2] % MOD;
                printf("%lld\n", ans);
                return 0;
        }
        int centroid = get[0];
        vector<int> sz(n);
        function<void (int, int)> get_size = [&](int u, int prev) {
                sz[u] = 1;
                for (auto v : g[u]) if (v != prev) {
                        get_size(v, u);
                        sz[u] += sz[v];
                }
        };
        get_size(centroid, -1);
        vector<int> subtrees;
        for (auto v : g[centroid]) subtrees.push_back(sz[v]);
        int k = (int) subtrees.size();
        vector<vector<long long>> dp(k + 1, vector<long long> (n + 1, 0));
        dp[0][1] = 1;
        for (int i = 0; i < k; i ++) {
                int num = subtrees[i];
                for (int j = 0; j <= n; j ++) {
                        for (int l = 0; l <= num; l ++) {
                                if (l & 1) {
                                        (dp[i + 1][j + num - l] -= dp[i][j] * C(num, l) % MOD * fact[num] % MOD * invfact[num - l] % MOD) %= MOD;
                                } else {
                                        (dp[i + 1][j + num - l] += dp[i][j] * C(num, l) % MOD * fact[num] % MOD * invfact[num - l] % MOD) %= MOD;
                                }
                        }
                }
        }
        long long ans = 0;
        for (int i = 0; i <= n; i ++) {
                (ans += dp[k][i] * fact[i] % MOD) %= MOD;
        }
        ans = ((ans % MOD) + MOD) % MOD;
        printf("%lld\n", ans);
        return 0;
}

### Prompt

Please create a solution in cpp 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 <algorithm>
#include <vector>
#include <functional>
#include <cassert>
#include <map>
#include <set>
using namespace std;

const int MOD = 1e9 + 7;

const int N = 5000001;
long long fact[N];
long long invfact[N];
long long inv[N];
void init() {
        fact[0] = fact[1] = 1;
        for (int i = 2; i < N; i ++) fact[i] = fact[i - 1] * i % MOD;
        inv[1] = 1;
        for (int i = 2; i < N; i ++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
        invfact[0] = invfact[1] = 1;
        for (int i = 2; i < N; i ++) invfact[i] = invfact[i - 1] * inv[i] % MOD;
}
long long C(long long n, long long r) {
        if (n < 0 || r < 0 || n < r) return 0;
        return fact[n] * invfact[n - r] % MOD * invfact[r] % MOD;
}

vector<int> get_centroids(vector<vector<int> > const & g, int root, set<int> const & forbidden) {
    map<int, int> available; {
        function<void (int, int)> go = [&](int i, int parent) {
            available.emplace(i, available.size());
            for (auto j : g[i]) if (j != parent and not forbidden.count(j)) {
                go(j, i);
            }
        };
        go(root, -1);
    }
    int n = available.size();
    vector<int> result;
    vector<int> size(n, -1);
    function<void (int, int)> go = [&](int x, int parent) {
        bool is_centroid = true;
        int i = available[x];
        size[i] = 1;
        for (auto y : g[x]) if (y != parent and !forbidden.count(y)) {
            int j = available[y];
            go(y, x);
            size[i] += size[j];
            if (size[j] > n / 2) is_centroid = false;
        }
        if (n - size[i] > n / 2) is_centroid = false;
        if (is_centroid) result.push_back(x);
    };
    go(root, -1);
    return result;
}

int main() {
        init();
        int n;
        scanf("%d", &n);
        vector<vector<int>> g(n);
        for (int i = 0; i < n - 1; i ++) {
                int a, b;
                scanf("%d%d", &a, &b);
                a --, b --;
                g[a].push_back(b);
                g[b].push_back(a);
        }
        set<int> forbidden;
        auto get = get_centroids(g, 0, forbidden);
        if (get.size() == 2) {
                long long ans = fact[n / 2] * fact[n / 2] % MOD;
                printf("%lld\n", ans);
                return 0;
        }
        int centroid = get[0];
        vector<int> sz(n);
        function<void (int, int)> get_size = [&](int u, int prev) {
                sz[u] = 1;
                for (auto v : g[u]) if (v != prev) {
                        get_size(v, u);
                        sz[u] += sz[v];
                }
        };
        get_size(centroid, -1);
        vector<int> subtrees;
        for (auto v : g[centroid]) subtrees.push_back(sz[v]);
        int k = (int) subtrees.size();
        vector<vector<long long>> dp(k + 1, vector<long long> (n + 1, 0));
        dp[0][1] = 1;
        for (int i = 0; i < k; i ++) {
                int num = subtrees[i];
                for (int j = 0; j <= n; j ++) {
                        for (int l = 0; l <= num; l ++) {
                                if (l & 1) {
                                        (dp[i + 1][j + num - l] -= dp[i][j] * C(num, l) % MOD * fact[num] % MOD * invfact[num - l] % MOD) %= MOD;
                                } else {
                                        (dp[i + 1][j + num - l] += dp[i][j] * C(num, l) % MOD * fact[num] % MOD * invfact[num - l] % MOD) %= MOD;
                                }
                        }
                }
        }
        long long ans = 0;
        for (int i = 0; i <= n; i ++) {
                (ans += dp[k][i] * fact[i] % MOD) %= MOD;
        }
        ans = ((ans % MOD) + MOD) % MOD;
        printf("%lld\n", ans);
        return 0;
}

```