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;
long long f[2][200500], g[2][200500], sf[2][200500], sg[2][200500], ans;
int a, b, k, t;
const long long MOD = 1000000007;
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  f[0][a + 100210] = 1;
  g[0][b + 100210] = 1;
  for (int j = -100000; j <= 100100; j++) {
    sf[0][j + 100210] = (sf[0][j + 100209] + f[0][j + 100210]) % MOD;
    sg[0][j + 100210] = (sg[0][j + 100209] + g[0][j + 100210]) % MOD;
  }
  for (int i = 1; i <= t; i++) {
    for (int j = -100000; j <= 100100; j++) {
      f[i & 1][j + 100210] =
          (sf[(i - 1) & 1][min(100210 + a + k * i, j + k + 100210)] -
           sf[(i - 1) & 1][max(100209 + a - k * i, j - k + 100209)]) %
          MOD;
      if (f[i & 1][j + 100210] < 0) f[i & 1][j + 100210] += MOD;
      sf[i & 1][j + 100210] =
          (sf[i & 1][j + 100209] + f[i & 1][j + 100210]) % MOD;
    }
    for (int j = -100000; j <= 100100; j++) {
      g[i & 1][j + 100210] =
          (sg[(i - 1) & 1][min(100210 + b + k * i, j + k + 100210)] -
           sg[(i - 1) & 1][max(100209 + b - k * i, j - k + 100209)]) %
          MOD;
      if (g[i & 1][j + 100210] < 0) g[i & 1][j + 100210] += MOD;
      sg[i & 1][j + 100210] =
          (sg[i & 1][j + 100209] + g[i & 1][j + 100210]) % MOD;
    }
  }
  for (int j = -100000; j <= 100100; j++) {
    ans = (ans + f[t & 1][j + 100210] * sg[t & 1][j + 100209]) % MOD;
  }
  printf("%lld\n", ans);
}
