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 + 5;
const int MOD = 1e9 + 7;
const int ze = 2e5;
long long dp[2][MAXN];
long long sum[MAXN];
int main() {
  int a, b, k, t;
  while (~scanf("%d %d %d %d", &a, &b, &k, &t)) {
    memset(dp, 0, sizeof(dp));
    int fl = 1;
    dp[1][ze] = 1;
    sum[0] = 0;
    for (int i = 1; i <= t; ++i) {
      for (int j = 1; j < MAXN; ++j) {
        sum[j] = (dp[fl][j] + sum[j - 1]) % MOD;
      }
      memset(dp[fl ^ 1], 0, sizeof(dp[fl ^ 1]));
      for (int j = 1; j < MAXN; ++j) {
        int t1 = max(0, j - k - 1);
        int t2 = min(MAXN - 1, j + k);
        dp[fl ^ 1][j] = (dp[fl ^ 1][j] + sum[t2] - sum[t1] + MOD) % MOD;
      }
      fl ^= 1;
    }
    for (int i = 1; i < MAXN; ++i) {
      sum[i] = (dp[fl][i] + sum[i - 1]) % MOD;
    }
    long long ans = 0;
    for (int i = 1; i < MAXN; ++i) {
      int tt = a - b + i;
      if (tt >= MAXN) tt = MAXN;
      if (tt < 1) tt = 1;
      ans = (ans + dp[fl][i] * sum[tt - 1] % MOD) % MOD;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
