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, sum;
long long lbound, ubound, size = 408000, zero = 204000;
vector<long long> cnt, nextCnt, pref;
const long long inf = 1e9 + 7;
int main() {
  cin >> a >> b >> k >> t;
  cnt.assign(size, 0);
  cnt[zero] = 1;
  lbound = zero;
  ubound = zero + 1;
  for (int i = 0; i < (2 * t); ++i) {
    lbound -= k;
    ubound += k;
    nextCnt.assign(size, 0);
    pref.clear();
    long long prefSum = 0;
    for (int j = 0; j < (k + 1); ++j) {
      pref.push_back(0);
    }
    for (int j = lbound; j < (ubound + 1); ++j) {
      prefSum = (prefSum + cnt[j]) % inf;
      pref.push_back(prefSum);
    }
    for (int j = 0; j < (k); ++j) {
      pref.push_back(prefSum);
    }
    int pl = 0, pr = 2 * k + 1;
    for (int j = lbound; j < (ubound); ++j) {
      nextCnt[j] = pref[pr] - pref[pl];
      if (nextCnt[j] < 0) {
        nextCnt[j] += inf;
      }
      pr++;
      pl++;
    }
    cnt = nextCnt;
  }
  sum = 0;
  for (int i = zero + b - a + 1; i < (size + 1); ++i) {
    sum = (sum + cnt[i]) % inf;
  }
  printf("%lld", sum);
}
