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 mod = 1e9 + 7;
long long sa, b, k, t;
const long long st = 2e5 + 1e3;
long long dp[500005];
long long a[500005], d[500005];
long long ra[500005], rd[500005];
long long ways[500005];
long long on = 1;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> sa >> b >> k >> t;
  dp[st] = 1;
  for (long long i = 0; i < t; i++) {
    memset(a, 0, sizeof(a));
    memset(d, 0, sizeof(d));
    memset(ra, 0, sizeof(ra));
    memset(rd, 0, sizeof(rd));
    for (long long j = 0; j < 500005; j++) {
      if (dp[j] == 0) continue;
      a[j] += dp[j] * (k + k + 1);
      d[j] -= dp[j];
      d[j + k + k + 1] += dp[j];
      ra[j - 1] += dp[j] * (k + k);
      rd[j - 1] -= dp[j];
      rd[j - k - k - 1] += dp[j];
    }
    memset(dp, 0, sizeof(dp));
    long long cur = 0;
    long long dif = 0;
    for (long long j = 0; j < 500005; j++) {
      cur += a[j];
      dp[j] += cur;
      dif += d[j];
      cur += dif;
    }
    cur = 0;
    dif = 0;
    for (long long j = 500005 - 1; j >= 0; j--) {
      cur += ra[j];
      dp[j] += cur;
      dif += rd[j];
      cur += dif;
    }
    for (long long i = 0; i < 500005; i++) {
      dp[i] %= mod;
      if (dp[i] < 0) dp[i] += mod;
    }
  }
  long long ans = 0;
  for (long long i = b - sa + 1 + st; i < 500005; i++) {
    ans += dp[i];
    if (ans >= mod) ans -= mod;
  }
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans;
  return 0;
}
