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 MAXN = 420000, p = 200000, mod = 1e9 + 7;
int d[2][MAXN], pref[2][MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  int cur = 1;
  d[0][a - b + p] = 1;
  for (int i = 0; i < MAXN; i++)
    pref[!cur][i] = (i == 0 ? 0 : pref[!cur][i - 1]) + d[!cur][i];
  for (int i = 1; i <= 2 * t; i++) {
    for (int j = 0; j < MAXN; j++) {
      d[cur][j] = ((pref[!cur][min(j + k, MAXN - 1)] -
                    (j - k > 0 ? pref[!cur][j - k - 1] : 0)) %
                       mod +
                   mod) %
                  mod;
      pref[cur][j] = ((j == 0 ? 0 : pref[cur][j - 1]) + d[cur][j]) % mod;
    }
    cur = !cur;
  }
  int ans = 0;
  cur = !cur;
  for (int j = p + 1; j < MAXN; j++) {
    ans = ((ans + d[cur][j]) % mod + mod) % mod;
  }
  cout << ans;
}
