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;
long long dp[2][101 * (1110 * 2) * 2];
long long ps[101 * (1110 * 2) * 2];
long long answ;
long long ml, mr, kb, md;
int main() {
  scanf("%I64d %I64d %I64d %I64d", &a, &b, &k, &t);
  long long cv = a - b;
  md = 2 * k * t, kb = 202001;
  ml = cv - md, mr = cv + md;
  dp[0 % 2][a - b + kb] = 1;
  for (register long long i = 1; i <= t; ++i) {
    long long s = 0, ls = 0, rs = 0;
    for (register long long ki = ml - 2 * k - 1; ki <= mr; ++ki) {
      long long cc = dp[(i - 1) % 2][min(mr + kb, ki + kb)],
                cn = dp[(i - 1) % 2][min(mr + kb, ki + 2 * k + kb)],
                cp = dp[(i - 1) % 2][ki - 2 * k - 1ll + kb];
      dp[i % 2][ki + kb] = s = (s - ls + rs + cn + 1000000007ll) % 1000000007ll;
      ls = (1000000007ll + ls - cp + cc) % 1000000007ll;
      rs = (1000000007ll + rs - cc + cn) % 1000000007ll;
    }
  }
  for (register long long i = 1; i <= mr; ++i)
    answ = (answ + dp[t % 2][i + kb]) % 1000000007ll;
  printf("%I64d\n", answ);
  return 0;
}
