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 K = 1000, T = 100, MXM = K * T + 110,
                MOD = 1000 * 1000 * 1000 + 7;
long long a, b, k, t, ans, dp[2][2 * MXM + 10], cp[2][2 * MXM + 10],
    ps[2][2 * MXM + 10], qs[2][2 * MXM + 10];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j;
  cin >> a >> b >> k >> t;
  dp[0][MXM + a] = 1;
  cp[0][MXM + b] = 1;
  for (i = MXM + min(a, b); i <= 2 * MXM; ++i) {
    ps[0][i] = ps[0][i - 1] + dp[0][i];
    qs[0][i] = qs[0][i - 1] + cp[0][i];
  }
  for (i = 1; i <= t; ++i)
    for (j = 1; j <= 2 * MXM; ++j) {
      dp[i % 2][j] = (MOD + ps[!(i % 2)][min(2 * MXM, j + k)] -
                      ps[!(i % 2)][max(0LL, j - k - 1)]) %
                     MOD;
      cp[i % 2][j] = (MOD + qs[!(i % 2)][min(2 * MXM, j + k)] -
                      qs[!(i % 2)][max(0LL, j - k - 1)]) %
                     MOD;
      ps[i % 2][j] = (ps[i % 2][j - 1] + dp[i % 2][j]) % MOD;
      qs[i % 2][j] = (qs[i % 2][j - 1] + cp[i % 2][j]) % MOD;
    }
  for (i = 1; i <= 2 * MXM; ++i)
    ans = (ans + ((dp[t % 2][i] * qs[t % 2][i - 1]) % MOD)) % MOD;
  cout << ans << '\n';
  return 0;
}
