Problem:
Score : 1600 points

Problem Statement

There are N zeros and M ones written on a blackboard.
Starting from this state, we will repeat the following operation: select K of the rational numbers written on the blackboard and erase them, then write a new number on the blackboard that is equal to the arithmetic mean of those K numbers.
Here, assume that N + M - 1 is divisible by K - 1.

Then, if we repeat this operation until it is no longer applicable, there will be eventually one rational number left on the blackboard.

Find the number of the different possible values taken by this rational number, modulo 10^9 + 7.

Constraints

1 ≦ N, M ≦ 2000

2 ≦ K ≦ 2000

N + M - 1 is divisible by K - 1.

Input

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

N M K

Output

Print the number of the different possible values taken by the rational number that will be eventually left on the blackboard, modulo 10^9 + 7.

Sample Input 1

2 2 2

Sample Output 1

5

There are five possible values for the number that will be eventually left on the blackboard: \frac{1}{4}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8} and \frac{3}{4}.

For example, \frac{3}{8} can be eventually left if we:

Erase 0 and 1, then write \frac{1}{2}.

Erase \frac{1}{2} and 1, then write \frac{3}{4}.

Erase 0 and \frac{3}{4}, then write \frac{3}{8}.

Sample Input 2

3 4 3

Sample Output 2

9

Sample Input 3

150 150 14

Sample Output 3

937426930

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

#include<bits/stdc++.h>
#define sqr(x) ((x)*(x))
#define ll long long
#define ull unsigned long long
#define ui unsigned int
#define ld long double
#define vi vector<int>
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > > 
#define pii pair<int,int> 
#define pb push_back
#define mp make_pair
#define debuge cerr<<"isok"<<endl
#define debug(x) cerr<<#x<<"="<<x<<endl
#define dprintf(...) fprintf(stderr,__VA_ARGS__)
#define SS second
#define FF first
#define ls (k<<1)
#define rs (k<<1|1)
#define clr(a,x) memset(a,x,sizeof(a))
#define cpy(a,x) memcpy(a,x,sizeof(a))
#define file(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
#define SZ(x) ((int)x.size())
using namespace std;
template<class T> inline void gmin(T &x,const T &y){if(x>y) x=y;}
template<class T> inline void gmax(T &x,const T &y){if(x<y) x=y;}
const int BufferSize=1<<16;
char buffer[BufferSize],*Bufferhead,*Buffertail;
bool Terminal;
inline char Getchar(){
	if(Bufferhead==Buffertail){
		int l=fread(buffer,1,BufferSize,stdin);
		if(!l){Terminal=1;return 0;}
		Buffertail=(Bufferhead=buffer)+l;
	}
	return *Bufferhead++;
}
template<class T>inline bool read(T &x){
	x=0;char c=Getchar(),rev=0;
	while(c<'0'||c>'9'){c=Getchar();rev|=c=='-';if(Terminal)return 0;}
	while(c>='0'&&c<='9') x=x*10+c-'0',c=Getchar();
	if(c=='.'){
		c=Getchar();double t=0.1;
		while(c>='0'&&c<='9') x=x+(c-'0')*t,c=Getchar(),t=t/10;
	}
	x=rev?-x:x;
	return 1;
}
template<class T1,class T2> inline bool read(T1 &x,T2 &y){return read(x)&read(y);}
template<class T1,class T2,class T3> inline bool read(T1 &x,T2 &y,T3 &z){return read(x)&read(y)&read(z);}
template<class T1,class T2,class T3,class T4> inline bool read(T1 &x,T2 &y,T3 &z,T4 &w){return read(x)&read(y)&read(z)&read(w);}
inline bool reads(char *x){
	char c=Getchar();
	while(c<33||c>126){c=Getchar();if(Terminal)return 0;}
	while(c>=33&&c<=126) (*x++)=c,c=Getchar();
	*x=0;return 1;
}
template<class T>inline void print(T x,const char c='\n'){
	if(!x){putchar('0');putchar(c);return;}
	if(x<0) putchar('-'),x=-x;
	int m=0,a[20];
	while(x) a[m++]=x%10,x/=10;
	while(m--) putchar(a[m]+'0');
	putchar(c);
}
//--------------------------------------------------head--------------------------------------------------

const int inf=0x3f3f3f3f;
const int N=4005,M=100005,mod=1e9+7;
template<class T,class S> inline void ch(T &x,const S y){x=(x+y)%mod;}
inline int exp(int x,int y,const int mod=::mod){
	int ans=1;
	while(y){
		if(y&1) ans=(ll)ans*x%mod;
		x=(ll)x*x%mod;y>>=1;
	}return ans;
}

int n,m,k,dp[2][N<<1],ans;

int main(){
#ifdef rqgao2014
	freopen("input.txt","r",stdin);
#endif
	read(n,m,k);
	int d=(n+m-1)/(k-1);
	debug(d);
	dp[0][0]=1;
	for(int i=1;i<=d;i++){
		for(int j=0;j<=n;j++) dp[i&1][j]=0;
		for(int j=0;j<=n;j++){
			int p=(n-j)%(k-1);if(!p) p+=k-1;
			if(!p||j+p>n||i*(k-1)-j-p+1>m) continue;
			ch(ans,dp[i-1&1][j]);
			for(int p=0;p<k;p++)
				ch(dp[i&1][j+p],dp[i-1&1][j]);
		}
	}
	print(ans);
	return 0;
}