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 = 2000000, md = (int)1e9 + 7;
int a, b, t, k, dp[2][N], g, delt[N];
inline void norm(int &x) {
  if (x < 0) x += md;
  if (x >= md) x -= md;
}
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  int n = 4 * k * t;
  g = 0;
  for (int i = 0; i <= n; i++) {
    dp[g][i] = dp[1 - g][i] = 0;
  }
  int sum = 0;
  dp[g][0] = 1;
  for (int i = 0; i < t << 1; i++) {
    for (int j = 0; j <= n; j++) {
      dp[1 - g][j] = 0;
      delt[j] = 0;
    }
    for (int j = 0; j <= n; j++) {
      delt[j] += dp[g][j];
      norm(delt[j]);
      delt[j + ((k << 1) + 1)] -= dp[g][j];
      norm(delt[j + ((k << 1) + 1)]);
    }
    for (int j = 0; j <= n; j++) {
      sum += delt[j];
      norm(sum);
      dp[1 - g][j] = sum;
    }
    g = 1 - g;
  }
  int ans = 0, dans = b - a + 2 * t * k;
  for (int i = max(dans + 1, 0); i <= n; i++) {
    ans += dp[g][i];
    norm(ans);
  }
  cout << ans << "\n";
  return 0;
}
