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 ara[3][5000000], mod = 1000000007;
long long bara[3][500000];
void fun2(int a, int t, int k, int low, int high) {
  int i, j, l, m, n, cnt = 1, prev, v;
  for (i = low; i <= high; i++, cnt++) {
    if (i == a) {
      break;
    }
  }
  for (i = cnt; i >= cnt - k; i--) {
    bara[1][i]++;
  }
  for (i = cnt + 1; i <= cnt + k; i++) {
    bara[1][i]++;
  }
  for (i = 1; i <= high; i++) {
    bara[1][i] = bara[1][i - 1] + bara[1][i];
  }
  return;
  for (j = 2; j <= t; j++) {
    v = j % 2;
    prev = (v + 1) % 2;
    for (i = 1; i <= high; i++) {
      l = i - k - 1;
      if (l < 0) l = 0;
      bara[v][i] = (bara[prev][i + k] % mod - bara[prev][l] % mod) % mod;
      if (bara[v][i] < 0) bara[v][i] = (bara[v][i] + mod) % mod;
      bara[v][i] = (bara[v][i] % mod + bara[v][i - 1] % mod) % mod;
    }
  }
}
int main() {
  int i, j, k, l, m, n, cnt = 1, prev, a, b, low, high, v, t, mal = 0;
  long long ans = 0;
  scanf("%d%d%d%d", &a, &b, &k, &t);
  low = -k * t;
  high = 200590;
  for (i = low; i <= high; i++, cnt++) {
    if (i == a) {
      break;
    }
  }
  for (i = cnt; i >= cnt - k; i--) {
    ara[1][i]++;
  }
  for (i = cnt + 1; i <= cnt + k; i++) {
    ara[1][i]++;
  }
  if (t > 1) {
    for (i = 1; i <= high; i++) ara[1][i] = ara[1][i] + ara[1][i - 1];
  }
  fun2(b, t, k, low, high);
  int goes = cnt + k + k;
  for (j = 2; j <= t; j++) {
    v = j % 2;
    prev = (v + 1) % 2;
    for (i = 1; i <= high; i++) {
      l = i - k - 1;
      if (l < 0) l = 0;
      int up = i + k;
      up = min(up, high);
      ara[v][i] = (ara[prev][up] - ara[prev][l]);
      if (ara[v][i] < 0) ara[v][i] = (ara[v][i] + mod) % mod;
      if (j < t) ara[v][i] = (ara[v][i] % mod + ara[v][i - 1] % mod) % mod;
      bara[v][i] = (bara[prev][up] - bara[prev][l]);
      if (bara[v][i] < 0) bara[v][i] = (bara[v][i] + mod) % mod;
      bara[v][i] = (bara[v][i] % mod + bara[v][i - 1] % mod) % mod;
    }
  }
  high = cnt + (k * t);
  v = t % 2;
  for (i = high; i >= 1; i--) {
    ans = (ans % mod + (ara[v][i] % mod * bara[v][i - 1] % mod) % mod) % mod;
  }
  printf("%I64d\n", ans);
}
