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 a, b, k, t, maxi, l, r, i, ans;
long long dp0[200005];
long long dp1[200005];
long long s[200005];
int main() {
  cin >> a >> b >> k >> t;
  maxi = k * t;
  dp1[maxi] = 1;
  while (t) {
    t--;
    for (i = 0; i <= 2 * maxi; i++) {
      dp0[i] = dp1[i];
      dp1[i] = 0;
    }
    for (i = 0; i <= 2 * maxi; i++) {
      l = i - k;
      if (l >= 0) {
        dp1[l] += dp0[i];
        if (dp1[l] > 1000000007) dp1[l] -= 1000000007;
      }
      r = i + k + 1;
      if (r <= 2 * maxi) {
        dp1[r] -= dp0[i];
        if (dp1[r] < 0) dp1[r] += 1000000007;
      }
    }
    for (i = 1; i <= 2 * maxi; i++) {
      dp1[i] += dp1[i - 1];
      if (dp1[i] > 1000000007) dp1[i] -= 1000000007;
    }
  }
  s[0] = dp1[0];
  for (i = 1; i <= 2 * maxi; i++) {
    s[i] = s[i - 1] + dp1[i];
    if (s[i] > 1000000007) s[i] -= 1000000007;
  }
  for (i = 0; i <= 2 * maxi; i++) {
    r = a - b + i - 1;
    if (r < 0) continue;
    if (r > 2 * maxi) r = 2 * maxi;
    ans += (s[r] * dp1[i]) % 1000000007;
    if (ans > 1000000007) ans -= 1000000007;
  }
  cout << ans;
  return 0;
}
