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;
vector<int> v, w;
int t, k, a, b;
int M = 1000000007;
int maximo(int p, int q) {
  if (p > q) return p;
  return q;
}
int main() {
  cin >> a >> b >> k >> t;
  int N = 4 * k * t + 2;
  v = w = vector<int>(N, 0);
  v[2 * k * t] = 1;
  for (int g = 0; g < t; g++) {
    for (int i = 0; i < N; i++) w[i] = 0;
    for (int i = k; i < N - k - 1; i++) {
      w[i - k] = (w[i - k] + v[i]) % M;
      w[i + k + 1] = (w[i + k + 1] - v[i]) % M;
    }
    for (int i = 1; i < N; i++) w[i] = (w[i] + w[i - 1]) % M;
    for (int i = 0; i < N; i++) v[i] = 0;
    for (int i = k; i < N - k - 1; i++) {
      v[i - k] = (v[i - k] + w[i]) % M;
      v[i + k + 1] = (v[i + k + 1] - w[i]) % M;
    }
    for (int i = 1; i < N; i++) v[i] = (v[i] + v[i - 1]) % M;
  }
  int solution = 0;
  for (int i = maximo(2 * k * t + b - a + 1, 0); i < N - 1; i++)
    solution = (solution + v[i]) % M;
  while (solution < 0) solution = solution + M;
  while (solution >= M) solution = solution - M;
  cout << solution << endl;
}
