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 MAX = 4e5 + 7;
const int le = 2e5;
const int mod = 1e9 + 7;
long long a, b, k, t;
long long dp[2][MAX];
long long sum[MAX];
int now = 0, pre = 1;
int main() {
  scanf("%lld%lld%lld%lld", &a, &b, &k, &t);
  dp[now][le] = 1;
  for (int i = 1; i <= t; i++) {
    for (int j = 1; j < MAX; j++) {
      sum[j] = dp[now][j] + sum[j - 1];
      sum[j] %= mod;
    }
    swap(now, pre);
    memset(dp[now], 0, sizeof(dp[now]));
    for (int j = 1; j < MAX; j++) {
      dp[now][j] += sum[min(MAX - 1LL, j + k)] - sum[max(0LL, j - k - 1)];
      dp[now][j] %= mod;
    }
  }
  for (int j = 1; j < MAX; j++) {
    sum[j] = dp[now][j] + sum[j - 1];
    sum[j] %= mod;
  }
  long long ans = 0;
  for (int j = 0; j < MAX; j++) {
    ans += sum[a + j - b - 1] % mod * dp[now][j] % mod;
    ans %= mod;
  }
  cout << (ans + mod) % mod << endl;
}
