Problem:
Score : 1500 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

Consider all integers between 1 and 2N, inclusive.
Snuke wants to divide these integers into N pairs such that:

Each integer between 1 and 2N is contained in exactly one of the pairs.

In exactly A pairs, the difference between the two integers is 1.

In exactly B pairs, the difference between the two integers is 2.

In exactly C pairs, the difference between the two integers is 3.

Note that the constraints guarantee that N = A + B + C, thus no pair can have the difference of 4 or more.

Compute the number of ways to do this, modulo 10^9+7.

Constraints

1 ≤ N ≤ 5000

0 ≤ A, B, C

A + B + C = N

Input

The input is given from Standard Input in the following format:

N A B C

Output

Print the answer.

Sample Input 1

3 1 2 0

Sample Output 1

2

There are two possibilities: 1-2, 3-5, 4-6 or 1-3, 2-4, 5-6.

Sample Input 2

600 100 200 300

Sample Output 2

522158867

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define REP(i,st,ed) for(register int i=st,i##end=ed;i<=i##end;++i)
#define DREP(i,st,ed) for(register int i=st,i##end=ed;i>=i##end;--i)
typedef long long ll;
template<typename T>inline bool chkmin(T &x,T y){return (y<x)?(x=y,1):0;}
template<typename T>inline bool chkmax(T &x,T y){return (y>x)?(x=y,1):0;}
inline int read(){
	int x;
	char c;
	int f=1;
	while((c=getchar())!='-' && (c>'9' || c<'0'));
	if(c=='-') f=-1,c=getchar();
	x=c^'0';
	while((c=getchar())>='0' && c<='9') x=(x<<1)+(x<<3)+(c^'0');
	return x*f;
}
inline ll readll(){
	ll x;
	char c;
	int f=1;
	while((c=getchar())!='-' && (c>'9' || c<'0'));
	if(c=='-') f=-1,c=getchar();
	x=c^'0';
	while((c=getchar())>='0' && c<='9') x=(x<<1ll)+(x<<3ll)+(c^'0');
	return x*f;
}
const int maxn=10000+10,mod=1e9+7;
int ksm(int x,int y){
	int res=1;
	while(y){
		if(y&1) res=(ll)res*x%mod;
		y>>=1;
		x=(ll)x*x%mod;
	}
	return res;
}
int fac[maxn],ifac[maxn];
inline void init(int n){
	fac[0]=1;
	REP(i,1,n) fac[i]=(ll)i*fac[i-1]%mod;
	ifac[n]=ksm(fac[n],mod-2);
	DREP(i,n,1) ifac[i-1]=(ll)i*ifac[i]%mod;
}
inline int calc(int S,int x4){
	if(S==0 && x4==0) return 1;
	if(x4==0) return 0;
	return (ll)fac[S+x4-1]*ifac[S]%mod*ifac[x4-1]%mod;
}
int main(){
	init(maxn-10);
	int n=read(),A=read(),B=read(),C=read(),ans=0;
	if(B&1) return printf("0\n"),0;
	REP(x1,0,A) REP(x3,0,C/3){
		int x2=A-x1,x4=B/2,S=C-x3*3-x2;
		if(S<0) continue;
		ans=(ans+(ll)fac[x1+x2+x3+x4]*ifac[x1]%mod*ifac[x2]%mod*ifac[x3]%mod*ifac[x4]%mod*calc(S,x4))%mod;
	}
	printf("%d\n",ans);
	return 0;
}