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>
const int P = 1000000007;
using namespace std;
int f[2][200005], sum[200005];
int main() {
  int a, b, k, t;
  scanf("%d%d%d%d", &a, &b, &k, &t);
  int i, j;
  f[0][k * t + 1] = 1;
  for (i = 1; i <= t; i++) {
    int cur = i & 1, pre = cur ^ 1;
    sum[0] = 0;
    for (j = 1; j <= t * k * 2 + 1; j++) sum[j] = (sum[j - 1] + f[pre][j]) % P;
    for (j = -k * i; j <= k * i; j++)
      f[cur][j + k * t + 1] = (sum[min(k * t * 2 + 1, j + k * t + 1 + k)] -
                               sum[max(0, j + k * t + 1 - k - 1)] + P) %
                              P;
  }
  sum[0] = 0;
  for (j = 1; j <= t * k * 2 + 1; j++) sum[j] = (sum[j - 1] + f[t & 1][j]) % P;
  int ans = 0;
  for (i = -k * t; i <= k * t; i++) {
    ans += 1LL * f[t & 1][i + k * t + 1] *
           ((sum[max(0, min(k * t * 2 + 1, i + a - b + k * t))] - sum[0] + P) %
            P) %
           P;
    if (ans > P) ans -= P;
  }
  printf("%d\n", ans);
  return 0;
}
