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;
inline void Add(int &x, int y) {
  if (y < 0) y += 1000000007;
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
int dp[105][100001 << 1], Sum[105][100001 << 1];
int main() {
  int a, b, K, t;
  scanf("%d%d%d%d", &a, &b, &K, &t);
  dp[0][100001] = 1;
  for (int i = (100001), i_ = ((100001 << 1) - 1); i <= i_; i++) Sum[0][i] = 1;
  for (int i = (1), i_ = (t); i <= i_; i++)
    for (int j = (1), j_ = ((100001 << 1) - 1); j <= j_; j++) {
      Add(dp[i][j], Sum[i - 1][min(j + K, (100001 << 1) - 1)]);
      Add(dp[i][j], -Sum[i - 1][max(j - K - 1, 0)]);
      Add(Sum[i][j], Sum[i][j - 1]);
      Add(Sum[i][j], dp[i][j]);
    }
  int ans = 0;
  a -= b + 1;
  for (int i = (1), i_ = ((100001 << 1) - 1); i <= i_; i++)
    Add(ans, 1LL * dp[t][i] *
                 (Sum[t][(100001 << 1) - 1] -
                  Sum[t][min(max(i - a - 1, 0), (100001 << 1) - 1)]) %
                 1000000007);
  printf("%d\n", ans);
  return 0;
}
