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;
int m[2][303300];
int *p1, *p2;
int offset = 151500;
int off = 120000;
int resa[303300];
int mod = 1e9 + 7;
void solve(int a, int k, int t) {
  p1 = m[0] + offset;
  p2 = m[1] + offset;
  memset(m, 0, sizeof(m));
  p1[a] = 1;
  for (int i = 0; i < t; i++) {
    if (i == 2) {
      bool f = 0;
    }
    int tmp = 0;
    memset(p2 - offset, 0, sizeof(m[0]));
    for (int j = -off; j < off; j++) {
      p2[j] = tmp;
      tmp = (((tmp - p1[j - k]) % mod + mod) % mod + p1[j + k + 1]) % mod;
    }
    swap(p1, p2);
  }
}
int main() {
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  solve(a, k, t);
  memcpy(resa, p1 - offset, sizeof(m[0]));
  solve(b, k, t);
  int res = 0;
  int tmp = 0;
  for (int i = -off; i < off; i++) {
    tmp = (tmp + p1[i]) % mod;
    res = (res + ((long long)resa[i + offset + 1] * tmp) % mod) % mod;
  }
  printf("%d\n", res);
  return 0;
}
