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 dp[105][200005];
long long x = 1e9 + 7, y;
int main() {
  long long a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q;
  scanf("%lld %lld %lld %lld", &a, &b, &c, &d);
  j = c * d;
  for (e = 0, f = 1; e <= j * 2; e++, f++) {
    if (e <= c * 2)
      dp[1][e] = f % x;
    else
      dp[1][e] = dp[1][e - 1];
  }
  for (e = 2; e <= d; e++) {
    for (f = 0; f <= j * 2; f++) {
      h = max(f - c * 2, 0LL);
      i = f;
      if (h - 1 < 0)
        g = 0;
      else
        g = dp[e - 1][h - 1];
      if (f == 0)
        k = 0;
      else
        k = dp[e][f - 1];
      dp[e][f] = (k + dp[e - 1][i] - g) % x;
    }
  }
  for (e = 0, o = 0; e <= j * 2; e++) {
    l = a + e - 1 - b;
    if (l < 0) continue;
    l = min(l, j * 2);
    if (e == 0)
      n = 0;
    else
      n = dp[d][e - 1];
    m = dp[d][l] * (dp[d][e] - n) % x;
    while (m < 0) m += x;
    while (m >= x) m -= x;
    o = (m + o) % x;
  }
  printf("%lld\n", o % x);
  return 0;
}
