Problem:
Score : 900 points

Problem Statement

Let N be an even number.

There is a tree with N vertices.
The vertices are numbered 1, 2, ..., N.
For each i (1 \leq i \leq N - 1), the i-th edge connects Vertex x_i and y_i.

Snuke would like to decorate the tree with ribbons, as follows.

First, he will divide the N vertices into N / 2 pairs.
Here, each vertex must belong to exactly one pair.
Then, for each pair (u, v), put a ribbon through all the edges contained in the shortest path between u and v.

Snuke is trying to divide the vertices into pairs so that the following condition is satisfied: "for every edge, there is at least one ribbon going through it."
How many ways are there to divide the vertices into pairs, satisfying this condition?
Find the count modulo 10^9 + 7.
Here, two ways to divide the vertices into pairs are considered different when there is a pair that is contained in one of the two ways but not in the other.

Constraints

N is an even number.

2 \leq N \leq 5000

1 \leq x_i, y_i \leq N

The given 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 divide the vertices into pairs, satisfying the condition, modulo 10^9 + 7.

Sample Input 1

4
1 2
2 3
3 4

Sample Output 1

2

There are three possible ways to divide the vertices into pairs, as shown below, and two satisfy the condition: the middle one and the right one.

Sample Input 2

4
1 2
1 3
1 4

Sample Output 2

3

There are three possible ways to divide the vertices into pairs, as shown below, and all of them satisfy the condition.

Sample Input 3

6
1 2
1 3
3 4
1 5
5 6

Sample Output 3

10

Sample Input 4

10
8 5
10 8
6 5
1 5
4 8
2 10
3 6
9 2
1 7

Sample Output 4

672

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
// #define int long long
// #define double long double
#define rep(i,a,b) for(int i=(int)(a); i<(int)(b); ++i)
#define repeq(i,a,b) for(int i=(int)(a); i<=(int)(b); ++i)
#define rrep(i,a,b) for(int i=(int)(a)-1; i>=(int)(b); --i)
#define rrepeq(i,a,b) for(int i=(int)(a); i>=(int)(b); --i)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
template<typename T> using Graph = vector<vector<T>>;
template<typename T> using Spacial = vector<vector<vector<T>>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename T> using greater_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const int MOD = 1e9+7;
const int MOD2 = 998244353;
// const double EPS = 1e-9;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
string interval[2] = {" ", "\n"}; // {" ", "\n"}

template<typename T> struct is_plural : false_type{};
template<typename T1, typename T2> struct is_plural<pair<T1, T2>> : true_type{};
template<typename T> struct is_plural<vector<T>> : true_type{};
template<typename T> struct is_plural<complex<T>> : true_type{};
template<> struct is_plural<string> : true_type{};

template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << " " << p.second; }
template<typename T> istream &operator>>(istream &is, vector<T> &vec) { for(auto itr = vec.begin(); itr != vec.end(); ++itr) is >> *itr; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { if(vec.empty()) return os; bool pl = is_plural<T>(); os << vec.front(); for(auto itr = ++vec.begin(); itr != vec.end(); ++itr) os << interval[pl] << *itr; return os; }

bool CoutYN(bool a, string y = "Yes", string n = "No") { cout << (a ? y : n) << "\n"; return a; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

long long modpow(int a, long long n, int mod = MOD) { long long ret = 1; do { if(n & 1) ret = ret * a % mod; a = 1LL * a * a % mod; } while(n >>= 1); return ret; }

template<typename T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
template<typename T> T LCM(T a, T b) { return a / GCD(a, b) * b; }

template<typename T1, typename T2> bool CompareBySecond(pair<T1, T2> a, pair<T1, T2> b) { return a.second != b.second ? a.second < b.second : a.first < b.first; }
template<typename T1, typename T2> bool CompareByInverse(pair<T1, T2> a, pair<T1, T2> b) { return a.first != b.first ? a.first < b.first : a.second > b.second; }


/* -------- <templates end> -------- */


template<uint_fast64_t Modulus = MOD>
struct Modint {
  using u64 = uint_fast64_t;
  u64 a;

  constexpr Modint(const u64 x = 0) noexcept : a(x % Modulus) {}

  constexpr Modint operator+(const Modint rhs) const noexcept {
    return Modint(*this) += rhs;
  }
  constexpr Modint operator-(const Modint rhs) const noexcept {
    return Modint(*this) -= rhs;
  }
  constexpr Modint operator*(const Modint rhs) const noexcept {
    return Modint(*this) *= rhs;
  }
  constexpr Modint operator/(const Modint rhs) const noexcept {
    return Modint(*this) /= rhs;
  }

  constexpr Modint &operator+=(const Modint rhs) noexcept {
    a += rhs.a;
    if (a >= Modulus) a -= Modulus;
    return *this;
  }
  constexpr Modint &operator-=(const Modint rhs) noexcept {
    if (a < rhs.a) a += Modulus;
    a -= rhs.a;
    return *this;
  }
  constexpr Modint &operator*=(const Modint rhs) noexcept {
    a = a * rhs.a % Modulus;
    return *this;
  }
  constexpr Modint &operator/=(Modint rhs) noexcept {
    u64 exp = Modulus - 2;
    while (exp) {
      if (exp & 1) *this *= rhs;
      rhs *= rhs;
      exp >>= 1;
    }
    return *this;
  }

  Modint pow(u64 t) const {
    if (!t) return 1;
    Modint x = pow(t>>1);
    x *= x;
    if (t&1) x *= *this;
    return x;
  }

  explicit operator bool() const {
    return a;
  }

  friend ostream &operator<<(ostream &os, const Modint &m) {
    return os << m.a;
  }
};

using mint = Modint<>;

void solve() {
  int n; cin >> n;
  Graph<int> g(n);
  rep(_,1,n) {
    int x, y; cin >> x >> y; --x, --y;
    g[x].emplace_back(y);
    g[y].emplace_back(x);
  }

  vector<mint> calc(n + 1);
  calc[0] = 1;
  for(int i=2; i<=n; i+=2) {
    calc[i] = calc[i-2] * (i-1);
  }

  Graph<mint> dp(n, {0, 1});

  auto dfs = [&](auto dfs, int v, int p) -> void {
    for(auto &nv : g[v]) {
      if(nv != p) {
        dfs(dfs, nv, v);

        auto cpy = dp[v];
        dp[v].assign(cpy.size() + dp[nv].size() - 1, 0);

        rep(j,0,cpy.size()) {
          dp[v][j] -= dp[nv][0] * cpy[j];
        }
        rep(i,1,dp[nv].size()) {
          rep(j,0,cpy.size()) {
            dp[v][i + j] += dp[nv][i] * cpy[j];
          }
        }
      }
    }

    for(int i=2; i<dp[v].size(); i+=2) {
      dp[v][0] += dp[v][i] * calc[i];
    }
  };

  dfs(dfs, 0, -1);
  cout << dp[0][0] << endl;
}


/* -------- <programs end> -------- */


signed main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  solve();
  return 0;
}