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 long long M = 1000000007;
const long long C = 2e5 + 1e4;
long long D[2 * C + 100];
long long D2[2 * C + 100];
long long a, b, k, t;
void refresh() {
  long long sum = 0;
  for (long long i = -C; i <= C; i++) {
    sum += D[C + i];
    sum %= M;
    if (C + i >= 2 * k + 1) {
      sum += M;
      sum -= D[C + i - 2 * k - 1];
    }
    sum %= M;
    if (C + i - k >= 0) {
      D2[C + i - k] = sum;
    }
  }
  for (long long i = -C; i <= C; i++) {
    D[C + i] = D2[C + i];
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b >> k >> t;
  t *= 2;
  D[C + a - b] = 1;
  while (t--) {
    refresh();
  }
  long long ans = 0;
  for (long long i = 1; i <= C; i++) {
    ans += D[C + i];
    ans %= M;
  }
  cout << ans;
}
