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 5.4.1) (judge verdict: Accepted)

//tree DP
//ABC036-D
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<time.h>
#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384

int max(int a,int b){if(a>b){return a;}return b;}
int min(int a,int b){if(a<b){return a;}return b;}
int zt(int a,int b){return max(a,b)-min(a,b);}
int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
int ceil(int a,int b){if(a%b==0){return a/b;}return (a/b)+1;}
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
int nHr(int a,int b){return nCr(a+b-1,b);}
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}
int sankaku(int x){return ((1+x)*x)/2;}
void swap(int *a,int *b){int c;c=(*a);(*a)=(*b);(*b)=c;}
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llnHr(long long a,long long b){return llnCr(a+b-1,b);}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long lldsum(long long x){long long r=0;while(x){r+=(x%10);x/=10;}return r;}
long long lldsumb(long long x,long long b){long long r=0;while(x){r+=(x%b);x/=b;}return r;}
long long llsankaku(long long x){return ((1+x)*x)/2;}
void llswap(long long *a,long long *b){long long c;c=(*a);(*a)=(*b);(*b)=c;}
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}
void dbswap(double *a,double *b){double c;c=(*a);(*a)=(*b);(*b)=c;}
void chswap(char *a,char *b){char c;c=(*a);(*a)=(*b);(*b)=c;}
int sortfncsj(const void *a,const void *b){if(*(int *)a>*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int sortfnckj(const void *a,const void *b){if(*(int *)a<*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int llsortfncsj(const void *a,const void *b){if(*(long long *)a>*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int llsortfnckj(const void *a,const void *b){if(*(long long *)a<*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int dbsortfncsj(const void *a,const void *b){if(*(double *)a>*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int dbsortfnckj(const void *a,const void *b){if(*(double *)a<*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int strsortfncsj(const void *a,const void *b){return strcmp((char *)a,(char *)b);}
int strsortfnckj(const void *a,const void *b){return strcmp((char *)b,(char *)a);}
int chsortfncsj(const void *a,const void *b){if(*(char *)a>*(char *)b){return 1;}if(*(char *)a==*(char *)b){return 0;}return -1;}
int chsortfnckj(const void *a,const void *b){if(*(char *)a<*(char *)b){return 1;}if(*(char *)a==*(char *)b){return 0;}return -1;}

typedef struct{
int val;
int node;
}sd;

int sdsortfnc(const void *a,const void *b){
if(((sd*)a)->val > ((sd*)b)->val){return -1;}
if(((sd*)a)->val < ((sd*)b)->val){return 1;}
return 0;
}

typedef struct{
    long long st;
    long long fi;
    long long kr;
}rs;

typedef struct{
    long long st;
    long long kz;
}mkj;

int sortfnc(const void *a,const void *b){
if(((rs*)a)->st > ((rs*)b)->st){return 1;}
if(((rs*)a)->st < ((rs*)b)->st){return -1;}
return 0;
}

void makemkj(rs g[],mkj x[],long long n){
    long long i,ms=0,nst=g[0].st;
    for(i=1;i<n;i++){
        if(g[i].st!=g[i-1].st){
            x[nst].kz=i-ms;
            x[nst].st=ms;
            nst=g[i].st;ms=i;
        }
    }
    x[nst].kz=n-ms;
    x[nst].st=ms;
}

long long dist[524288],par[524288];
void dfs(long long t,long long l,long long bp,rs g[],mkj x[]){
  long long i;
  if(dist[t]<=l){return;}
  dist[t]=l;
  par[t]=bp;
  for(i=x[t].st;i<x[t].st+x[t].kz;i++){
    dfs(g[i].fi,l+1,t,g,x);
  }
}

int main(void){
  long long i,j,n,m,k,a,b,c,h,w,r=0,l,t,p,q;
  long long bp,bq;
  long long ***dp=malloc(sizeof(long long**)*5005);
  for(i=0;i<5005;i++){
    dp[i]=malloc(sizeof(long long*)*5005);
    for(j=0;j<5005;j++){
      dp[i][j]=malloc(sizeof(long long)*2);
      dp[i][j][0]=0;
      dp[i][j][1]=0;
    }
  }
  long long ndp[5005][2]={0};
  long long wei[524288]={0};
  rs g[524288];
  mkj x[524288];
  sd dat[524288];
  long long f[8192]={0};
  f[0]=1;
  for(i=2;i<8192;i+=2){
    f[i]=((i-1)*f[i-2])%mod;
  }
  scanf("%lld",&n);
  for(i=0;i<(n-1);i++){
    scanf("%lld%lld",&a,&b);
    g[2*i].st=a;
    g[2*i].fi=b;
    g[2*i].kr=1;
    g[2*i+1].st=b;
    g[2*i+1].fi=a;
    g[2*i+1].kr=1;
  }
  qsort(g,2*(n-1),sizeof(g[0]),sortfnc);
  makemkj(g,x,2*(n-1));
  for(i=0;i<=n;i++){
    dist[i]=inf;
  }
  dfs(1,0,-1,g,x);
  for(i=0;i<n;i++){
    dat[i].node=i+1;
    dat[i].val=dist[i+1];
  }
  qsort(dat,n,sizeof(dat[0]),sdsortfnc);
  for(i=0;i<n;i++){
    w=dat[i].node;
    dp[w][1][0]=1;
    wei[w]++;
    for(j=x[w].st;j<x[w].st+x[w].kz;j++){
      if(par[w]==g[j].fi){continue;}
      for(p=1;p<=wei[w];p++){
        for(q=1;q<=wei[g[j].fi];q++){
          for(bp=0;bp<2;bp++){
            for(bq=0;bq<2;bq++){
              t=(bp+bq+1)%2;
              k=dp[g[j].fi][q][bq];
              k*=f[q];k%=mod;
              k*=dp[w][p][bp];k%=mod;
              ndp[p][t]=(ndp[p][t]+k)%mod;
              t=(bp+bq)%2;
              k=dp[g[j].fi][q][bq];
              k*=dp[w][p][bp];k%=mod;
              ndp[p+q][t]=(ndp[p+q][t]+k)%mod;
            }
          }
        }
      }
      wei[w]+=wei[g[j].fi];
      for(p=1;p<=wei[w];p++){
        for(q=0;q<2;q++){
          dp[w][p][q]=ndp[p][q];
          ndp[p][q]=0;
        }
      }
    }
  }
  for(i=2;i<=n;i+=2){
    r=(r+dp[1][i][0]*f[i])%mod;
    r=(r+mod-((dp[1][i][1]*f[i])%mod))%mod;
  }
  printf("%lld\n",r);
  return 0;
}