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;
template <typename A, typename B>
inline void chmin(A &a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b) {
  if (a < b) a = b;
}
const long long mod = 1000000007;
long long dp[2][666666];
inline void add(long long &a, long long b) {
  a += b;
  if (a >= b) a -= mod;
  if (a < 0) a += mod;
}
signed main() {
  long long A, B, K, T;
  cin >> A >> B >> K >> T;
  dp[0][A - B + 300000] = 1;
  for (long long i = 0; i < 2 * T; i++) {
    long long sum = 0;
    for (long long j = 0; j < 2 * K + 1; j++) add(sum, dp[i & 1][j]);
    for (long long j = K; j + K + 1 < 666666; j++) {
      dp[(i + 1) & 1][j] = sum;
      add(sum, dp[i & 1][j - K]);
      add(sum, -dp[i & 1][j + K + 1]);
    }
  }
  long long ans = 0;
  for (long long i = 300000 + 1; i < 666666; i++) add(ans, dp[0][i]);
  cout << ans << endl;
  return 0;
}
