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 D {

  static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  
  final static long MOD = 1_000_000_007L;

  public static void main(String[] args) throws IOException {

    String[] line = br.readLine().split(" ");
    int a = Integer.valueOf(line[0]);
    int b = Integer.valueOf(line[1]);
    final int K = Integer.valueOf(line[2]);
    final int T = Integer.valueOf(line[3]) + 1; 

    // Initialize constants
    final int MAX = (K * T * 2) + 1;
    final int MID = MAX / 2;

    // Pre-compute by keeping rolling sum of length (K*2 + 1) of row above
    long[][] arr = new long[T][MAX];
    arr[0][MID] = 1;
    for (int t = 1; t < T; t++) {

      // Initialize rolling sum
      long sum = 0;
      for (int i = 0; i <= K && i < MAX; i++)
        sum = (sum + arr[t-1][i]) % MOD;
      
      for (int i = 0; i < MAX; i++) {

        arr[t][i] = sum;

        int indexToRemove = i - K;
        if (indexToRemove >= 0)
          sum -= arr[t-1][indexToRemove];
        int indexToAdd = i + K + 1;
        if (indexToAdd < MAX)
          sum += arr[t-1][indexToAdd];
        sum = ((sum % MOD) + MOD) % MOD;

      }

    }

    // Compute partial sums of last row
    long[] sums = new long[MAX];
    sums[0] = arr[T-1][0];
    for (int i = 1; i < MAX; i++)
      sums[i] = (sums[i-1] + arr[T-1][i]) % MOD;

    long nGames = 0;
    int pointsMemoryMustWinBy = b - a + 1;
    for (int i = 0; i < MAX; i++) {

      int memoryIndex = i;
      int highestLexaIndex = memoryIndex - pointsMemoryMustWinBy;

      // System.out.println(memoryIndex + " " + highestLexaIndex);
      if (highestLexaIndex >= 0) {
        if (highestLexaIndex >= MAX)
          nGames = (nGames + (arr[T-1][memoryIndex] * sums[MAX - 1])) % MOD;
        else
          nGames = (nGames + (arr[T-1][memoryIndex] * sums[highestLexaIndex])) % MOD;
      }

    }

    System.out.println(nGames);
  
  }

}