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 maxn = 4e5 + 10;
const long long mod = 1e9 + 7;
int a, b, k, t, mx;
long long dp1[2][maxn], dp2[2][maxn], cur1, cur, ans, sm[2][maxn];
int main() {
  scanf("%d%d%d%d", &a, &b, &k, &t);
  cur = 0;
  mx = max(2 * k * t + a, 2 * k * t + b);
  dp1[cur][a + k * t] = 1;
  for (int i = 1; i <= mx; i++) sm[cur][i] = dp1[cur][i] + sm[cur][i - 1];
  for (int i = 1; i <= t; i++) {
    for (int j = 0; j <= mx; j++) {
      dp1[cur ^ 1][j] =
          ((sm[cur][min(mx, j + k)] - sm[cur][max(0, j - k - 1)]) % mod + mod) %
          mod;
      sm[cur ^ 1][j] = dp1[cur ^ 1][j];
      if (j) sm[cur ^ 1][j] = (sm[cur ^ 1][j] + sm[cur ^ 1][j - 1]) % mod;
    }
    memset(dp1[cur], 0, sizeof(dp1[cur]));
    memset(sm[cur], 0, sizeof(sm[cur]));
    cur ^= 1;
  }
  cur1 = cur;
  cur = 0;
  dp2[cur][b + k * t] = 1;
  memset(sm[cur], 0, sizeof(sm[cur]));
  for (int i = 1; i <= mx; i++) sm[cur][i] = dp2[cur][i] + sm[cur][i - 1];
  for (int i = 1; i <= t; i++) {
    for (int j = 0; j <= mx; j++) {
      dp2[cur ^ 1][j] =
          ((sm[cur][min(mx, j + k)] - sm[cur][max(0, j - k - 1)]) % mod + mod) %
          mod;
      sm[cur ^ 1][j] = dp2[cur ^ 1][j];
      if (j) sm[cur ^ 1][j] = (sm[cur ^ 1][j] + sm[cur ^ 1][j - 1]) % mod;
    }
    memset(dp2[cur], 0, sizeof(dp2[cur]));
    memset(sm[cur], 0, sizeof(sm[cur]));
    cur ^= 1;
  }
  for (int i = 1; i <= mx; i++)
    ans = (ans + dp1[cur1][i] * sm[cur][i - 1] % mod) % mod;
  printf("%lld\n", ans);
  return 0;
}
