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>
const int MAXT = 100;
const int MAXK = 1000;
const int MOD = 1000000007;
const int ZERO = 100100;
const int INF = 200200;
using namespace std;
long long f[MAXT + 10][MAXT * MAXK * 2 + 300], ans = 0;
int a, b, t, k;
void input(int &r) {
  r = 0;
  char t = getchar();
  while (!isdigit(t)) t = getchar();
  while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
}
int main() {
  input(a);
  input(b);
  input(k);
  input(t);
  int i, j;
  f[0][ZERO] = 1;
  for (i = 1; i <= t; i++) {
    for (j = 1; j <= INF; j++)
      f[i - 1][j] = (f[i - 1][j] + f[i - 1][j - 1]) % MOD;
    for (j = 0; j <= INF; j++)
      f[i][j] =
          ((f[i - 1][min(INF, j + k)] - f[i - 1][max(j - k - 1, 0)]) + MOD) %
          MOD;
  }
  for (j = 1; j <= INF; j++) f[t][j] = (f[t][j] + f[t][j - 1]) % MOD;
  for (j = 0; j <= INF; j++) {
    int tt = min(INF, j + a - b - 1);
    if (tt < 0)
      continue;
    else {
      long long temp = f[t][j] - f[t][j - 1];
      ans = (ans + (f[t][j] - f[t][j - 1] + MOD) * f[t][tt]) % MOD;
    }
  }
  printf("%I64d\n", ans);
}
