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 const MIN = -10000000000001;
long long const MAX = 10000000000001;
int const N = 200000;
long long const MOD = 10e8 + 7;
vector<int> numbers(200000, 0);
vector<int> tmp;
vector<vector<int> > graph(N, tmp);
int main() {
  long long a, b, k, t;
  cin >> a >> b >> k >> t;
  vector<long long> dp(300000, 0);
  vector<long long> cur(300000, 0);
  dp[a] = 1;
  long long maxler = (2 * k + 1) * t + max(a, b);
  for (int i = 0; i < t; i++) {
    long long sum = 0;
    for (int j = 0; j < 2 * k + 1; j++) {
      sum = (sum + dp[j]) % MOD;
      cur[j] = sum;
    }
    for (int j = 2 * k + 1; j < maxler; j++) {
      if (sum + dp[j] - dp[j - 2 * k - 1] < 0)
        sum = MOD + (sum + dp[j] - dp[j - 2 * k - 1]);
      else
        sum = (sum + dp[j] - dp[j - 2 * k - 1]) % MOD;
      cur[j] = sum;
    }
    dp = cur;
  }
  vector<long long> dpb(300000, 0);
  dpb[b] = 1;
  for (int i = 0; i < t; i++) {
    long long sum = 0;
    for (int j = 0; j < 2 * k + 1; j++) {
      sum = (sum + dpb[j]) % MOD;
      cur[j] = sum;
    }
    for (int j = 2 * k + 1; j < maxler; j++) {
      if (sum + dpb[j] - dpb[j - 2 * k - 1] < 0)
        sum = MOD + (sum + dpb[j] - dpb[j - 2 * k - 1]);
      else
        sum = (sum + dpb[j] - dpb[j - 2 * k - 1]) % MOD;
      cur[j] = sum;
    }
    dpb = cur;
  }
  long long sum = 0;
  for (int i = 0; i < a; i++) {
    sum = (sum + dpb[i]) % MOD;
  }
  long long count = 0;
  for (int i = a; i < maxler; i++) {
    count = (count + (dp[i] * sum) % MOD) % MOD;
    sum = (sum + dpb[i]) % MOD;
  }
  sum = 0;
  cout << count << endl;
  return 0;
}
