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 a, b, k, t;
const long long MOD = 1000000007;
long long mi(long long x, long long n = MOD - 2) {
  long long y = 1;
  while (n) {
    if (n & 1) y = (y * x) % MOD;
    x = (x * x) % MOD;
    n >>= 1;
  }
  return y;
}
long long fc[300123], ifc[300123];
long long nck(long long nn, long long kk) {
  if (nn < 0 or kk < 0 or kk > nn) return 0;
  long long ret = fc[nn];
  ret = (ret * ifc[kk]) % MOD;
  ret = (ret * ifc[nn - kk]) % MOD;
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> a >> b >> k >> t;
  fc[0] = ifc[0] = 1;
  for (long long i = 1; i <= 300000; i++) {
    fc[i] = (i * fc[i - 1]) % MOD;
    ifc[i] = mi(fc[i]);
  }
  long long tar = (2 * k * t + b - a), ans = 0;
  for (long long r = 0; r <= 2 * t; r++) {
    long long lt = nck(2 * t, r);
    long long t2 = tar - (2 * k + 1) * r;
    long long rt = nck(2 * t + t2, t2);
    long long term = (lt * rt) % MOD;
    if (r % 2)
      ans = (ans - term + MOD) % MOD;
    else
      ans = (ans + term) % MOD;
  }
  long long tot = mi(2 * k + 1, 2 * t);
  ans = (tot + MOD - ans) % MOD;
  cout << ans << '\n';
  return 0;
};
