712_D. Memory and Scores
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.
```python
mod=10**9+7
f=[0]*500000

def POW(a,b):
	if(b==0):
		return 1
	if(b&1):
		return POW(a,b//2)**2*a%mod
	else:
		return POW(a,b//2)**2

def C(n,m):
	if(m>n):
		return 0
	t=f[n]*POW(f[m],mod-2)%mod*POW(f[n-m],mod-2)%mod
	return t


f[0]=1
for i in range(1,500000):
	f[i]=f[i-1]*i%mod
a,b,k,t=map(int,input().split(' '))

ans=0
for i in range(0,2*t+1):
	t1=POW(-1,i)*C(2*t,i)%mod
	t2=(C(210000+2*k*t-a+b+2*t-1-(2*k+1)*i+1,2*t)-C(1+2*k*t-a+b+2*t-1-(2*k+1)*i,2*t))%mod
	ans=(ans+t1*t2)%mod
print(ans)

```