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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class MemoryScores {
    
    static final int MOD = (int) (1e9 + 7);
    
    void solve() {
        int a = in.nextInt(), b = in.nextInt(), k = in.nextInt(), t = in.nextInt();
        
        long[][] dp = new long[t + 1][2 * t * k + 1];
        
        dp[0][t * k] = 1;
        for (int i = 1; i <= t; i++) {
            long sum = 0;
            for (int j = 0; j <= k; j++) {
                sum = (sum + dp[i - 1][j]) % MOD;
            }
            for (int v = 0; v <= 2 * t * k; v++) {
                dp[i][v] = sum;
                if (v + k + 1 <= 2 * t * k) sum = (sum + dp[i - 1][v + k + 1]) % MOD;
                if (v - k >= 0) sum = (sum - dp[i - 1][v - k] + MOD) % MOD;
            }
        }
        
        long[] cnt = new long[2 * t * k + 1];
        cnt[0] = dp[t][0];
        for (int i = 1; i <= 2 * t * k; i++) cnt[i] = (cnt[i - 1] + dp[t][i]) % MOD;
        
        long ans = 0;
        for (int v = 0; v <= 2 * t * k; v++) {
            int w = v + a - b - 1;
            if (w < 0) continue;
            w = Math.min(w, 2 * t * k);
            ans = (ans + cnt[w] * dp[t][v] % MOD) % MOD;
        }
        out.println(ans);
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new MemoryScores().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
