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 N = 6e5 + 5;
const int offset = 3e5;
const int MOD = 1e9 + 7;
int a, b, k, t;
int dp[2][2][N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> a >> b >> k >> t;
  for (int player = 0; player <= 1; player++) {
    for (int diff = -offset; diff <= offset; diff++) {
      dp[player][t + 1 & 1][diff + offset] = (diff > 0);
      dp[player][t + 1 & 1][diff + offset] +=
          dp[player][t + 1 & 1][diff + offset - 1];
      dp[player][t + 1 & 1][diff + offset] %= MOD;
    }
  }
  for (int turn = t; turn >= 1; turn--) {
    for (int diff = -offset; diff <= offset; diff++) {
      int player = 1;
      int &ans = dp[player][turn & 1][diff + offset];
      ans = dp[player ^ 1][turn + 1 & 1][diff + k + offset] -
            dp[player ^ 1][turn + 1 & 1][diff - k - 1 + offset];
      ans += MOD;
      ans %= MOD;
      ans += dp[player][turn & 1][diff - 1 + offset];
      ans %= MOD;
    }
    for (int diff = -offset; diff <= offset; diff++) {
      int player = 0;
      int &ans = dp[player][turn & 1][diff + offset];
      ans = dp[player ^ 1][turn & 1][diff + k + offset] -
            dp[player ^ 1][turn & 1][diff - k - 1 + offset];
      ans += MOD;
      ans %= MOD;
      ans += dp[player][turn & 1][diff - 1 + offset];
      ans %= MOD;
    }
  }
  int ans = dp[0][1][a - b + offset] - dp[0][1][a - b - 1 + offset];
  ans += MOD;
  ans %= MOD;
  cout << ans;
  return 0;
}
