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 long long N = 1005;
const long long LMT = 2 * N * N;
long long dp[LMT];
long long sum[LMT];
const long long MOD = 1e9 + 7;
void add(long long &a, long long b) {
  a += b;
  a %= MOD;
}
void init_dp(long long k, long long t) {
  memset(dp, 0, sizeof(dp));
  dp[k * t] = 1;
  long long lmt = 2 * k * t;
  for (long long i = 1; i <= t; i++) {
    memset(sum, 0, sizeof(sum));
    sum[0] = dp[0];
    for (long long j = 1; j <= lmt; j++) {
      add(sum[j], sum[j - 1] + dp[j]);
    }
    for (long long j = lmt; j >= 0; j--) {
      long long l = max(1LL, j - k);
      long long r = min(lmt, j + k);
      dp[j] = 0;
      add(dp[j], sum[r] - sum[l - 1] + MOD);
    }
  }
}
long long got(long long a, long long b, long long vl) {
  if (vl < a) {
    return 0;
  } else if (vl > b) {
    return sum[b];
  } else {
    return sum[vl];
  }
}
long long solve(long long k, long long t, long long a, long long b) {
  long long res = 0;
  long long lmt = k * t * 2;
  sum[0] = dp[0];
  for (long long i = 1; i <= lmt; i++) {
    sum[i] = 0;
    add(sum[i], sum[i - 1] + dp[i]);
  }
  for (long long i = 0; i <= lmt; i++) {
    add(res, dp[i] * got(0, lmt, i + a - b - 1) % MOD);
  }
  return res % MOD;
}
int main() {
  long long k, t, a, b;
  cin >> a >> b >> k >> t;
  init_dp(k, t);
  cout << solve(k, t, a, b) << endl;
  return 0;
}
