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;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
const int N = 3e5 + 505;
const int MOD = 1e9 + 7;
int shift = 100000;
long long pr[N];
long long dp[2][N];
long long calc(int i, int j) {
  if (j < 0) return 0;
  return dp[i][j];
}
int main() {
  int a, b, k, t;
  cin >> a >> b >> k >> t;
  dp[0][0 + shift] = 1;
  int now = 0;
  for (int i = 1; i <= t; i++) {
    now = !now;
    for (int j = -shift; j <= shift; j++) {
      dp[now][j + shift] = calc(now, j + shift - 1) +
                           calc(!now, j + k + shift) -
                           calc(!now, j - k - 1 + shift);
      dp[now][j + shift] = (dp[now][j + shift] % MOD + MOD) % MOD;
    }
  }
  for (int i = shift; i >= -shift; i--) {
    pr[i + shift] = pr[i + 1 + shift] + dp[now][i + shift];
    pr[i + shift] %= MOD;
  }
  long long ans = 0;
  for (int i = -shift; i <= shift; i++) {
    int o = i + b - a + 1;
    ans += (dp[now][i + shift] * pr[max(0, o + shift)]) % MOD;
    ans %= MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
