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 mod = 1e9 + 7;
const int ori = 200200;
long long dp[2][400405];
int main() {
  int a, b, k, t;
  scanf("%d%d%d%d", &a, &b, &k, &t);
  int now = 1, nex = 0, from = ori + a - b, to = ori + a - b;
  dp[now][ori + a - b] = 1;
  for (int i = 1; i <= 2 * t; ++i) {
    memset(dp[nex], 0, sizeof dp[nex]);
    int nex_from = from - k, nex_to = to + k;
    for (int j = nex_from, l = 1; l <= 2 * k + 1; ++j, ++l) {
      dp[nex][j] = dp[now][from + l - 1];
    }
    for (int j = nex_from + 2 * k + 1, l = 2 * k + 1; j <= ori + a - b;
         ++j, ++l) {
      dp[nex][j] = dp[now][from + l] - dp[now][from + l - 2 * k - 1];
      if (dp[nex][j] < 0) dp[nex][j] += mod;
    }
    for (int j = nex_from, l = 1; j <= ori + a - b; ++j, ++l) {
      dp[nex][j] = dp[nex][j] + dp[nex][j - 1];
      if (dp[nex][j] > mod) dp[nex][j] -= mod;
      dp[nex][2 * (ori + a - b) - j] = dp[nex][j];
    }
    from = nex_from, to = nex_to, swap(now, nex);
  }
  long long ans = 0;
  for (int i = ori + 1; i <= to; ++i) {
    ans += dp[now][i];
    if (ans > mod) ans -= mod;
  }
  printf("%lld\n", ans);
  return 0;
}
