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 = 1150 * 1150, L = 2 * N - 1, M = 1e9 + 7;
int a, b, k, t;
int dp[4][N * 2];
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  dp[0][a + N] = 1;
  for (int i = 1, I = 1; i <= t; ++i, I ^= 1) {
    int j = max(0, N + a - k * i);
    int J = min(L - 1, N + a + k * i);
    int left = max(0, j - k), right = min(L - 1, j + k);
    int sum = 0;
    while (left <= right) (sum += dp[I ^ 1][left++]) %= M;
    while (j <= J) {
      dp[I][j] = sum;
      if (j - k >= 0) (sum -= dp[I ^ 1][j - k]) %= M;
      ++j;
      if (j + k < L) (sum += dp[I ^ 1][j + k]) %= M;
    }
  }
  dp[2][b + N] = 1;
  for (int i = 1, I = 3; i <= t; ++i, I ^= 1) {
    int j = max(0, N + b - k * i);
    int J = min(L - 1, N + b + k * i);
    int left = max(0, j - k), right = min(L - 1, j + k);
    int sum = 0;
    for (int i = left; i <= right; ++i) (sum += dp[I ^ 1][left++]) %= M;
    while (j <= J) {
      dp[I][j] = sum;
      if (j - k >= 0) (sum -= dp[I ^ 1][j - k]) %= M;
      ++j;
      if (j + k < L) (sum += dp[I ^ 1][j + k]) %= M;
    }
  }
  t &= 1;
  for (int i = 1; i < L; ++i) (dp[t | 2][i] += dp[t | 2][i - 1]) %= M;
  int ans = 0;
  for (int i = 1; i < L; ++i) {
    (ans += 1ll * dp[t][i] * dp[t | 2][i - 1] % M) %= M;
  }
  printf("%d\n", (ans + M) % M);
}
