Imagine that you are in a building that has exactly n floors. You can move between the floors in a lift. Let's number the floors from bottom to top with integers from 1 to n. Now you're on the floor number a. You are very bored, so you want to take the lift. Floor number b has a secret lab, the entry is forbidden. However, you already are in the mood and decide to make k consecutive trips in the lift.

Let us suppose that at the moment you are on the floor number x (initially, you were on floor a). For another trip between floors you choose some floor with number y (y ≠ x) and the lift travels to this floor. As you cannot visit floor b with the secret lab, you decided that the distance from the current floor x to the chosen y must be strictly less than the distance from the current floor x to floor b with the secret lab. Formally, it means that the following inequation must fulfill: |x - y| < |x - b|. After the lift successfully transports you to floor y, you write down number y in your notepad.

Your task is to find the number of distinct number sequences that you could have written in the notebook as the result of k trips in the lift. As the sought number of trips can be rather large, find the remainder after dividing the number by 1000000007 (109 + 7).

Input

The first line of the input contains four space-separated integers n, a, b, k (2 ≤ n ≤ 5000, 1 ≤ k ≤ 5000, 1 ≤ a, b ≤ n, a ≠ b).

Output

Print a single integer — the remainder after dividing the sought number of sequences by 1000000007 (109 + 7).

Examples

Input

5 2 4 1


Output

2


Input

5 2 4 2


Output

2


Input

5 3 4 1


Output

0

Note

Two sequences p1, p2, ..., pk and q1, q2, ..., qk are distinct, if there is such integer j (1 ≤ j ≤ k), that pj ≠ qj.

Notes to the samples:

  1. In the first sample after the first trip you are either on floor 1, or on floor 3, because |1 - 2| < |2 - 4| and |3 - 2| < |2 - 4|. 
  2. In the second sample there are two possible sequences: (1, 2); (1, 3). You cannot choose floor 3 for the first trip because in this case no floor can be the floor for the second trip. 
  3. In the third sample there are no sought sequences, because you cannot choose the floor for the first trip.

Source: codeforces | Difficulty: HARD
Tags: Dynamic programming

Combinatorics


Solutions:
def solve(n, st, k):
	MOD = int(1000000000.0 + 7)
	dp = [0] * (n + 1)
	prefix_sum = [0] * (n + 1)
	dp[st] = 1
	for times in range(k):
		prefix_sum[0] = 0
		for i in range(1, n + 1):
			prefix_sum[i] = prefix_sum[i - 1] + dp[i]
			if prefix_sum[i] >= MOD:
				prefix_sum[i] -= MOD
		for i in range(1, n + 1):
			dp[i] = prefix_sum[n] - prefix_sum[i] + prefix_sum[i - 1] - prefix_sum[i >> 1]
			while dp[i] < 0:
				dp[i] += MOD
			while dp[i] >= MOD:
				dp[i] -= MOD
	return sum(dp) % MOD

def main():
	(n, a, b, k) = [int(i) for i in input().split()]
	if a > b:
		print(solve(n - b, a - b, k))
	else:
		print(solve(b - 1, b - a, k))
main()


(n, a, b, k) = map(int, input().split())
dp = [[0 for i in range(n + 2)] for j in range(2)]
dp[0][a] = 1
now = 1
last = 0
mod = 1000000007
for i in range(k):
	for j in range(1, n + 1):
		d = max(abs(j - b) - 1, 0)
		if j != n:
			dp[now][j + 1] = (dp[last][j] + dp[now][j + 1]) % mod
			dp[now][min(j + d + 1, n + 1)] = (dp[now][min(j + d + 1, n + 1)] - dp[last][j]) % mod
		if j != 1:
			dp[now][j] = (dp[now][j] - dp[last][j]) % mod
			dp[now][max(j - d, 1)] = (dp[now][max(j - d, 1)] + dp[last][j]) % mod
	for i1 in range(1, n + 2):
		dp[now][i1] = (dp[now][i1] + dp[now][i1 - 1]) % mod
		dp[last][i1] = 0
	aux = now
	now = last
	last = aux
print(sum(dp[last]) % mod)


import io
import os
import sys
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def printd(*args, **kwargs):
	pass

def get_str():
	return input().decode().strip()

def rint():
	return map(int, input().split())

def oint():
	return int(input())
mod = 1000000007
(n, a, b, k) = rint()
if a > b:
	(a, b) = (n - a + 1, n - b + 1)
a -= 1
b -= 1
printd(n, a, b, k)
d = [0] * n
d[a] = 1
ps = [0] * b
ps[0] = d[0]
for j in range(1, b):
	ps[j] = ps[j - 1] + d[j]
	while ps[j] > mod:
		ps[j] -= mod
printd(n, a, b, k)
printd(d, ps)
for i in range(k):
	for j in range(b):
		if (b + j) % 2:
			t = (b + j) // 2
		else:
			t = (b + j) // 2 - 1
		if j == 0:
			d[j] = ps[t] - ps[j]
		else:
			d[j] = ps[t] - ps[j] + ps[j - 1]
		while d[j] > mod:
			d[j] -= mod
		while d[j] < 0:
			d[j] += mod
	ps[0] = d[0]
	for j in range(1, b):
		ps[j] = ps[j - 1] + d[j]
		while ps[j] > mod:
			ps[j] -= mod
		while ps[j] < 0:
			ps[j] += mod
		printd(d, ps)
ans = ps[b - 1]
print(ans % mod)


from sys import stdin

def parser():
	return map(int, stdin.readline().split())
mod = pow(10, 9) + 7
(n, a, b, k) = parser()
if a > b:
	a = n - a + 1
	n = n - b
	b = n + 1
else:
	n = b - 1
prefix_sum = [0 for x in range(n + 1)]
secuences = [0 for x in range(n + 1)]
secuences[a] = 1
for i in range(k):
	prefix_sum[0] = secuences[0]
	for j in range(1, n + 1):
		prefix_sum[j] = prefix_sum[j - 1] + secuences[j]
		prefix_sum[j] %= mod
	for j in range(1, n + 1):
		distance = b - j
		mid_distance = 0
		if distance % 2 == 0:
			mid_distance = distance // 2 - 1
		else:
			mid_distance = distance // 2
		secuences[j] = prefix_sum[j - 1] + prefix_sum[j + mid_distance] - prefix_sum[j]
		secuences[j] %= mod
print(sum(secuences) % mod)


import collections
import time
import os
import sys
import bisect
import heapq
from typing import List

def solve(N, A, B, K):
	MOD = 1000000007
	dp = [0 for _ in range(N + 1)]
	dp[A] = 1
	for k in range(1, K + 1):
		ndp = [0 for _ in range(N + 1)]
		for x in range(1, N + 1):
			d = abs(x - B)
			if d <= 1:
				continue
			(l, r) = (max(x - d + 1, 1), min(x + d, N + 1))
			if l < x:
				ndp[l] = (ndp[l] + dp[x]) % MOD
				ndp[x] = (ndp[x] - dp[x]) % MOD
			if x < r:
				if x + 1 <= N:
					ndp[x + 1] = (ndp[x + 1] + dp[x]) % MOD
				if r <= N:
					ndp[r] = (ndp[r] - dp[x]) % MOD
		v = 0
		for x in range(N + 1):
			v += ndp[x]
			v %= MOD
			ndp[x] = v
		dp = ndp
	return sum(dp) % MOD
(N, A, B, K) = map(int, input().split())
print(solve(N, A, B, K))


from sys import stdin
MOD = int(1000000000.0) + 7

def solve(tc):
	(n, a, b, k) = map(int, stdin.readline().split())
	dp = [1 for i in range(n + 1)]
	dp[b] = 0
	prefix = [1 for i in range(n + 1)]
	for i in range(k):
		for j in range(1, n + 1):
			prefix[j] = prefix[j - 1] + dp[j]
			prefix[j] %= MOD
		for j in range(1, b - 1):
			dist = b - j
			start = max(0, j - dist)
			end = min(b - 1, j + dist - 1)
			dp[j] = prefix[j - 1] - prefix[start]
			if dp[j] < 0:
				dp[j] += MOD
			dp[j] %= MOD
			dp[j] += prefix[end] - prefix[j]
			if dp[j] < 0:
				dp[j] += MOD
			dp[j] %= MOD
		dp[b - 1] = dp[b] = 0
		if b + 1 <= n:
			dp[b + 1] = 0
		for j in range(b + 2, n + 1):
			dist = j - b
			start = max(b, j - dist)
			end = min(n, j + dist - 1)
			dp[j] = prefix[j - 1] - prefix[start]
			if dp[j] < 0:
				dp[j] += MOD
			dp[j] %= MOD
			dp[j] += prefix[end] - prefix[j]
			if dp[j] < 0:
				dp[j] += MOD
			dp[j] %= MOD
	print(dp[a])
tc = 1
solve(tc)