# これ解くのに 998244353 時間かかった！
# ぼくが弱いだけで, ★3.0 …かも？

# よく考えると T が大きければ何もする必要はないし, 時を戻すときも N+M-1 以上時を戻す必要もない.
# そう思うと, 時間の初期値を200として, 時間を400までもつ.
# -> 時間は 0 より戻れないし, 400 より進めない. でもそれでいい.
# その間で, (時間, 位置) の拡張 dijkstra 法を行うとよい.
# O((N+M)NM log(NM)) になる.

import heapq

n, m, k, t = map(int,input().split())
ikeru = [[[] for j in range(m)] for i in range(n)]
for i in range(n):
	for j in range(m):
		for x, y in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
			if 0 <= i + x < n and 0 <= j + y < m:
				ikeru[i][j].append((i + x, j + y))

modoru = [[(-1,-1) for j in range(m)] for i in range(n)]
for i in range(k):
	a, b, c, d = map(int,input().split())
	a -= 1
	b -= 1
	modoru[a][b] = (c, d)

tmax = 401

d = [10 ** 18] * n * m * tmax
d[200] = 0

pq = [(-d[200], 200)]

r1 = m * tmax

while pq:
	tmp, nowv = heapq.heappop(pq)
	fnv = nowv
	nowt = nowv % tmax
	nowv //= tmax
	b = nowv % m
	a = nowv // m
	tmp = -tmp
	if tmp > d[fnv]: continue
	for x, y in ikeru[a][b]:
		dist = tmp
		targ = x*r1 + y*tmax + min(nowt+1, 400)
		if dist < d[targ]:
			d[targ] = dist
			heapq.heappush(pq, (-dist, targ))
	if modoru[a][b][0] == -1: continue
	c, dd = modoru[a][b]
	dist = tmp + dd
	targ = a*r1 + b*tmax + max(0,nowt-c)
	if dist < d[targ]:
		d[targ] = dist
		heapq.heappush(pq, (-dist, targ))

ans = 10 ** 18
for ts in range(401):
	if ts - 200 > t: continue
	ans = min(ans, d[(n-1)*r1 + (m-1)*tmax + ts])

if ans == 10 ** 18:
	print(-1)
else:
	print(ans)