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;
int dp[2][400110];
int main() {
  int a, b, K, n;
  scanf("%d%d%d%d", &a, &b, &K, &n);
  int i, j, f = 0;
  dp[0][0] = 1;
  for (i = 1; i <= n; i++) {
    f ^= 1;
    memset(dp[f], 0, sizeof(dp[f]));
    for (j = 1; j < 400110; j++) {
      dp[!f][j] = (dp[!f][j] + dp[!f][j - 1]) % 1000000007;
    }
    int sum = 0;
    for (j = 0; j < 400110; j++) {
      sum = (sum + dp[!f][j]) % 1000000007;
      if (j >= 2 * K + 1) {
        sum = (sum - dp[!f][j - (2 * K + 1)] * 2LL) % 1000000007;
      }
      if (j >= 4 * K + 2) {
        sum = (sum + dp[!f][j - (4 * K + 2)]) % 1000000007;
      }
      dp[f][j] = sum;
    }
  }
  int Ans = 0;
  for (i = 0; i < 400110; i++) {
    if (a - b + i - 2 * K * n > 0) {
      Ans = (Ans + dp[f][i]) % 1000000007;
    }
  }
  if (Ans < 0) {
    Ans += 1000000007;
  }
  printf("%d\n", Ans);
  return 0;
}
