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 = 1000000007;
const int DELTA = 100001;
const int N = 200001;
int a, b, k, t, ans;
int f[300010], d[300010], s[300010];
int main() {
  cin >> a >> b >> k >> t;
  f[0 + DELTA] = 1;
  while (t--) {
    for (int i = 1; i <= N; i++) d[i] = 0;
    for (int i = 1; i <= N; i++)
      if (f[i])
        d[i - k] = (d[i - k] + f[i]) % MOD,
              d[i + k + 1] = (d[i + k + 1] - f[i] + MOD) % MOD;
    for (int i = 1; i <= N; i++) f[i] = (f[i - 1] + d[i]) % MOD;
  }
  for (int i = 1; i <= N; i++) s[i] = (s[i - 1] + f[i]) % MOD;
  for (int i = -100000; i <= 100000; i++)
    if (a + i - b - 1 + DELTA <= N && a + i - b - 1 + DELTA >= 0)
      ans = (ans + ((long long)f[i + DELTA] * s[a + i - b - 1 + DELTA]) % MOD) %
            MOD;
    else if (a + i - b - 1 + DELTA > N)
      ans = (ans + ((long long)f[i + DELTA] * s[N]) % MOD) % MOD;
  cout << ans << endl;
  return 0;
}
