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>
using namespace std;
int p[2][200100];
int main() {
  int a, b, i, imax, j, k, res, s, t, w;
  cin >> a >> b >> k >> t;
  k *= 2;
  for (i = 0; i <= k; i++) p[0][i] = 1;
  t--;
  s = 0;
  imax = k;
  while (t-- > 0) {
    imax += k;
    s = 1 - s;
    w = 0;
    for (i = 0; i <= k; i++) {
      w = (w + p[1 - s][i]) % 1000000007;
      p[s][i] = w;
    }
    for (i = k + 1; i <= imax; i++) {
      w = (1ll * w + p[1 - s][i] - p[1 - s][i - k - 1] + 1000000007) %
          1000000007;
      p[s][i] = w;
    }
  }
  res = 0;
  w = 0;
  for (j = 0; j < a - b; j++) w = (w + p[s][j]) % 1000000007;
  for (i = max(0, b - a); i <= imax; i++) {
    res = (1ll * p[s][i] * w + res) % 1000000007;
    w = (w + p[s][j++]) % 1000000007;
  }
  cout << res << endl;
  return 0;
}
