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 pos;
long long dp[2][400345], sum[2][400345];
int main() {
  int a, b, t, i, j, k;
  scanf("%d %d %d %d", &a, &b, &k, &t);
  pos = 101 + (t * (2 * k));
  int lim = ((4 * k + 1) * t) + 211;
  dp[0][a - b + pos] = 1LL;
  sum[0][0] = 0LL;
  for (i = 1; i <= lim; i++) sum[0][i] = dp[0][i] + sum[0][i - 1];
  int c = 1;
  for (i = 1; i <= t; i++) {
    long long an = 0LL, p = 2 * k + 1LL;
    for (j = 0; j <= 2 * k; j++, p--)
      an = (an + (p * dp[!c][j])) % 1000000007LL;
    dp[c][0] = an;
    for (j = 1; j <= lim; j++) {
      long long pre = dp[c][j - 1];
      long long suff = sum[!c][min(lim, j + 2 * k)] - sum[!c][j - 1];
      long long pref = sum[!c][j - 1] - sum[!c][max(0, j - 2 * k - 2)];
      dp[c][j] = (pre + suff - pref) % 1000000007LL;
      dp[c][j] += 1000000007LL;
      dp[c][j] %= 1000000007LL;
    }
    sum[c][0] = dp[c][0];
    for (j = 1; j <= lim; j++)
      sum[c][j] = (dp[c][j] + sum[c][j - 1]) % 1000000007LL;
    c = !c;
  }
  long long ans = 0LL;
  for (j = pos + 1; j <= lim; j++) ans = (ans + dp[!c][j]) % 1000000007LL;
  printf("%lld\n", ans);
  return 0;
}
