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>
int main() {
  int a, b, k, t;
  long long ta[2][200300] = {};
  long long tb[2][200300] = {};
  long long w, r(0);
  std::scanf("%d%d%d%d", &a, &b, &k, &t);
  ta[0][(200300 / 2) + a] = tb[0][(200300 / 2) + b] = 1;
  for (int h(0); h < t; ++h) {
    std::memset(ta[h & 1 ^ 1], 0, sizeof(ta[0]));
    for (int i((200300 / 2) + a - h * k); i <= (200300 / 2) + a + h * k; ++i) {
      ta[h & 1 ^ 1][i - k] += ta[h & 1][i];
      ta[h & 1 ^ 1][i + k + 1] -= ta[h & 1][i];
    }
    for (int i((200300 / 2) + a - (h + 1) * k + 1);
         i <= (200300 / 2) + a + (h + 1) * k + 1; ++i)
      ta[h & 1 ^ 1][i] += ta[h & 1 ^ 1][i - 1];
    for (int i((200300 / 2) + a - (h + 1) * k);
         i <= (200300 / 2) + a + (h + 1) * k; ++i)
      ta[h & 1 ^ 1][i] %= 1000000007;
  }
  for (int h(0); h < t; ++h) {
    std::memset(tb[h & 1 ^ 1], 0, sizeof(tb[0]));
    for (int i((200300 / 2) + b - h * k); i <= (200300 / 2) + b + h * k; ++i) {
      tb[h & 1 ^ 1][i - k] += tb[h & 1][i];
      tb[h & 1 ^ 1][i + k + 1] -= tb[h & 1][i];
    }
    for (int i((200300 / 2) + b - (h + 1) * k + 1);
         i <= (200300 / 2) + b + (h + 1) * k + 1; ++i)
      tb[h & 1 ^ 1][i] += tb[h & 1 ^ 1][i - 1];
    for (int i((200300 / 2) + b - (h + 1) * k);
         i <= (200300 / 2) + b + (h + 1) * k; ++i)
      tb[h & 1 ^ 1][i] %= 1000000007;
  }
  for (int i(1); i < 200300; ++i) {
    ta[t & 1][i] += ta[t & 1][i - 1];
    if (1000000007 <= ta[t & 1][i]) ta[t & 1][i] -= 1000000007;
  }
  for (int i(0); i < 200300; ++i) {
    w = tb[t & 1][i] * (1000000007 + ta[t & 1][200300 - 1] - ta[t & 1][i]) %
        1000000007;
    r += w;
    if (1000000007 <= r) r -= 1000000007;
  }
  std::printf("%d\n", (int)r);
  return 0;
}
