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 int N = (int)3e5 + 10;
const int base = N / 2;
const long long mod = (long long)1e9 + 7;
int a, b, k, t;
long long p[N][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> a >> b >> k >> t;
  int c = 0;
  for (int i = base - k; i <= base + k; ++i) p[i][c] = 1;
  for (int i = base - k; i < N; ++i) p[i][c] += p[i - 1][c];
  for (int i = 1; i <= t - 1; ++i) {
    for (int j = base - k * (i + 1); j <= base + k * (i + 1); ++j) {
      p[j][c ^ 1] = (p[j + k][c] - p[j - k - 1][c] + mod) % mod;
    }
    for (int j = base - k * (i + 1); j < N; ++j)
      (p[j][c ^ 1] += p[j - 1][c ^ 1]) %= mod;
    for (int j = base - k * i; j < N; j++) p[j][c] = 0;
    c ^= 1;
  }
  long long ans = 0LL;
  for (int i = base - k * t; i <= base + k * t; ++i) {
    int new_a = a + (i - base);
    int j = new_a - b - 1 + base;
    ans = (ans + ((p[i][c] - p[i - 1][c]) * p[j][c]) % mod) % mod;
  }
  cout << ans << endl;
  return 0;
}
