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;
const int mod = 1e9 + 7;
const int N = 300010;
inline int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline int mult(int a, int b) { return (a * (long long)b) % mod; }
int dp[2][N];
int a, b, k, t;
int sum[N];
void solve(int st, int c) {
  memset(sum, 0, sizeof sum);
  for (int i = st; i < N; i++) {
    sum[i] = 1;
  }
  for (int j = 0; j < t; j++) {
    for (int i = 0; i < N; i++) {
      int s = i - k - 1;
      int t = i + k;
      if (t >= N) t = N - 1;
      if (s < 0)
        dp[c][i] = sum[t];
      else
        dp[c][i] = (sum[t] - sum[s] + mod) % mod;
    }
    sum[0] = dp[c][0];
    for (int i = 1; i < N; i++) {
      sum[i] = add(dp[c][i], sum[i - 1]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> a >> b >> k >> t;
  solve(N / 2 + a, 0);
  solve(N / 2 + b, 1);
  for (int i = N - 2; i >= 0; i--) {
    if (dp[0][i] == 0 && i <= a) break;
    dp[0][i] = add(dp[0][i], dp[0][i + 1]);
  }
  int res = 0;
  for (int i = 0; i + 1 < N; i++) {
    res = add(res, mult(dp[1][i], dp[0][i + 1]));
  }
  cout << res << "\n";
  return 0;
}
