Memory and his friend Lexa are competing to get higher score in one popular computer game. Memory starts with score a and Lexa starts with score b. In a single turn, both Memory and Lexa get some integer in the range [ - k;k] (i.e. one integer among  - k, - k + 1, - k + 2, ..., - 2, - 1, 0, 1, 2, ..., k - 1, k) and add them to their current scores. The game has exactly t turns. Memory and Lexa, however, are not good at this game, so they both always get a random integer at their turn.

Memory wonders how many possible games exist such that he ends with a strictly higher score than Lexa. Two games are considered to be different if in at least one turn at least one player gets different score. There are (2k + 1)2t games in total. Since the answer can be very large, you should print it modulo 109 + 7. Please solve this problem for Memory.

Input

The first and only line of input contains the four integers a, b, k, and t (1 ≤ a, b ≤ 100, 1 ≤ k ≤ 1000, 1 ≤ t ≤ 100) — the amount Memory and Lexa start with, the number k, and the number of turns respectively.

Output

Print the number of possible games satisfying the conditions modulo 1 000 000 007 (109 + 7) in one line.

Examples

Input

1 2 2 1


Output

6


Input

1 1 1 2


Output

31


Input

2 12 3 1


Output

0

Note

In the first sample test, Memory starts with 1 and Lexa starts with 2. If Lexa picks  - 2, Memory can pick 0, 1, or 2 to win. If Lexa picks  - 1, Memory can pick 1 or 2 to win. If Lexa picks 0, Memory can pick 2 to win. If Lexa picks 1 or 2, Memory cannot win. Thus, there are 3 + 2 + 1 = 6 possible games in which Memory wins.
import java.util.*;
import java.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=1000000007;
	long gcd(long a,long b) {
		return a==0?b:b>=a?gcd(b%a,a):gcd(b,a);
	}
	void work() {
		int a=in.nextInt(),b=in.nextInt(),k=in.nextInt(),t=in.nextInt();
		long[] dp1=new long[250000];
		long[] dp2=new long[250000];
		dp1[120000]=1;
		dp2[b-a+120000]=1;
		for(int p=0;p<t;p++) {
			long[] ndp1=new long[250000];
			long[] ndp2=new long[250000];
			for(int i=1000;i<240000;i++) {
				if(dp1[i]>0) {
					ndp1[i-k]=(ndp1[i-k]+dp1[i]+mod)%mod;
					ndp1[i+k+1]=(ndp1[i+k+1]-dp1[i]+mod)%mod;
				}
				if(dp2[i]>0) {
					ndp2[i-k]=(ndp2[i-k]+dp2[i]+mod)%mod;
					ndp2[i+k+1]=(ndp2[i+k+1]-dp2[i]+mod)%mod;
				}
			}
			for(int i=1;i<250000;i++) {
				ndp1[i]=(ndp1[i]+ndp1[i-1]+mod)%mod;
				ndp2[i]=(ndp2[i]+ndp2[i-1]+mod)%mod;
			}
			dp1=ndp1;
			dp2=ndp2;
		}
		long ret=0;
		long sum=0;
		for(int i=0;i<250000;i++) {
			ret=(ret+(dp1[i]*sum))%mod;
			sum=(sum+dp2[i])%mod;
		}
		out.println(ret);
	}
}	



class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}
	
	
	public String next() 
	{
		if(st==null || !st.hasMoreElements())
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	public int nextInt() 
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}