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 MOD = 1000000007;
const int H = 200010;
const int N = 2 * H + 1;
int a, b, k, t;
int dp[110][N + 10], pre[N + 10];
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  a += H;
  b += H;
  dp[t][H] = 1;
  for (int i = 0; i <= N; i++) {
    pre[i] = (long long)((i == 0 ? 0 : pre[i - 1]) + dp[t][i]) % MOD;
  }
  for (int idx = t - 1; idx >= 0; idx--) {
    for (int rem = N; rem >= 0; rem--) {
      int r = min(N, rem + k);
      int l = max(0, rem - k);
      dp[idx][rem] =
          (long long)(dp[idx][rem] + pre[r] - (l == 0 ? 0 : pre[l - 1])) % MOD;
    }
    for (int i = 0; i <= N; i++) {
      pre[i] = (long long)(pre[i - 1] + dp[idx][i]) % MOD;
    }
  }
  long long ways = 0;
  for (int i = 0; i <= N; i++) {
    pre[i] = (long long)((i == 0 ? 0 : pre[i - 1]) + dp[0][i]) % MOD;
  }
  int diff = b - a + 1;
  for (int i = -H; i <= H; i++) {
    int l = -H;
    int r = i - diff;
    if (r < l) continue;
    l += H;
    r += H;
    ways = (ways + (long long)((pre[r] - (l == 0 ? 0 : pre[l - 1])) + MOD) %
                       MOD * dp[0][i + H]) %
           MOD;
  }
  printf("%lld\n", ways);
  return 0;
}
