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 T = 100 + 10;
const int KT = 2 * 1000 * 101 + 10;
const int ZR = 2 * 1000 * 101 + 10;
const int M = 1e9 + 7;
void pl(int &a, int b) {
  a += b;
  if (a >= M) a -= M;
  if (a < 0) a += M;
}
int ans[2 * KT];
int pref[2 * KT];
void CalcPref() {
  for (int j = 0; j < 2 * KT; ++j) {
    pref[j] = 0;
    if (j > 0) pl(pref[j], pref[j - 1]);
    pl(pref[j], ans[j]);
  }
}
int GetS(int l, int r) {
  l = max(l, 0);
  r = min(r, 2 * KT - 1);
  if (l > r) return 0;
  int ret = pref[r];
  if (l - 1 >= 0) pl(ret, -pref[l - 1]);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  ans[ZR + a - b] = 1;
  for (int i = 1; i <= t; ++i) {
    CalcPref();
    int cur = 0;
    for (int j = 0; j < 2 * KT; ++j) {
      pl(cur, -GetS(j - (2 * k + 1), j - 1));
      pl(cur, GetS(j, j + 2 * k));
      ans[j] = cur;
    }
  }
  CalcPref();
  cout << GetS(ZR + 1, 2 * KT - 1) << '\n';
}
