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.
#include <bits/stdc++.h>
int dp[2][2 * 200100 + 5];
int main() {
  int a, b, k, t;
  scanf("%d %d %d %d", &a, &b, &k, &t);
  int delta = -200100;
  dp[0][a - b - delta] = 1;
  for (int i = a - b - delta; i <= 2 * 200100; ++i) {
    dp[0][i] = 1;
  }
  for (int i = 1; i <= t; ++i) {
    int v = 2 * k + 1;
    for (int z = 0; z <= 2 * k; ++z) {
      dp[i % 2][0] = (dp[i % 2][0] + v * dp[(i + 1) % 2][z]) % 1000000007;
      --v;
    }
    for (int j = 1; j <= 2 * 200100; ++j) {
      if (j - 2 >= 0) {
        dp[i % 2][j] =
            (1000000007 + dp[i % 2][j - 1] - dp[i % 2][j - 2]) % 1000000007;
      } else {
        dp[i % 2][j] = (1000000007 + dp[i % 2][j - 1]) % 1000000007;
      }
      dp[i % 2][j] =
          (1000000007 + dp[i % 2][j] - dp[(i + 1) % 2][j - 1]) % 1000000007;
      if (j - 2 * k - 2 >= 0) {
        dp[i % 2][j] =
            (dp[i % 2][j] + dp[(i + 1) % 2][j - 2 * k - 2]) % 1000000007;
      }
      if (j + 2 * k <= 2 * 200100) {
        dp[i % 2][j] = (dp[i % 2][j] + dp[(i + 1) % 2][j + 2 * k]) % 1000000007;
      } else {
        dp[i % 2][j] =
            (dp[i % 2][j] + dp[(i + 1) % 2][2 * 200100]) % 1000000007;
      }
      dp[i % 2][j] =
          (1000000007 + dp[i % 2][j] - dp[(i + 1) % 2][j - 1]) % 1000000007;
      dp[i % 2][j] = (dp[i % 2][j - 1] + dp[i % 2][j]) % 1000000007;
    }
  }
  int sol =
      (1000000007 + dp[t % 2][2 * 200100] - dp[t % 2][2 * 200100 + delta]) %
      1000000007;
  printf("%d", sol);
  return 0;
}
