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;
const int32_t MOD = 1e9 + 7;
const int32_t N = 1e5 + 3000;
int32_t a, b, k, t;
int64_t dem = 0, d = 0;
int64_t f[2 * N], s[2 * N];
int main() {
  cin >> a >> b >> k >> t;
  f[N] = 1;
  for (int32_t i = 1; i <= t; i++) {
    for (int32_t j = -k * i; j <= k * i; j++)
      s[j + N] = (s[j - 1 + N] + f[j + N]) % MOD;
    for (int32_t j = -k * i; j <= k * i; j++) {
      int32_t x = min(k + j, k * i);
      int32_t y = max(j - k, -k * i) - 1;
      f[j + N] = (s[x + N] - s[y + N] + MOD) % MOD;
    }
  }
  for (int32_t i = b - a - k * t; i <= k * t; i++) {
    d = (d + f[N + i + a - b - 1]) % MOD;
    dem = (dem + d * f[N + i]) % MOD;
  }
  cout << dem;
}
