"""

N,M が小さいんだよな
つまり、Tが大きい場合は無視で構わない

後は…

dp[x][y][T] = x,y に時刻Tで居る為に必要な最小コスト
で良いか

時刻は, T-(N-1-M-1)

-200 ～ T 位まで考慮しておけばいいかな

"""

import sys
from sys import stdin
import heapq

N,M,K,T = map(int,stdin.readline().split())

AB_to_CD = [[None] * M for i in range(N)]

for i in range(K):
    A,B,C,D = map(int,stdin.readline().split())
    A -= 1
    B -= 1
    AB_to_CD[A][B] = (C,D)

dp = [[[float("inf")] * 500 for i in range(M)] for j in range(N)]

dp[0][0][0] = 0

q = [ (0,0,0,0) ]

while q:

    cost,x,y,t = heapq.heappop(q)
    
    if (x,y) == (N-1,M-1) and t <= T:
        print (cost)
        sys.exit()

    if dp[x][y][t] != cost:
        continue

    if t <= N+M+10:

        for nx,ny in ( (x+1,y),(x-1,y),(x,y-1),(x,y+1) ):
            if 0 <= nx < N and 0 <= ny < M and dp[nx][ny][t+1] > dp[x][y][t]:
                dp[nx][ny][t+1] = dp[x][y][t]
                heapq.heappush( q,(dp[nx][ny][t+1] , nx , ny , t+1) )

    if AB_to_CD[x][y] != None:
        C,D = AB_to_CD[x][y]

        nexT = max(-200 , t-C+1)
        if dp[x][y][nexT] > D + dp[x][y][t]:
            dp[x][y][nexT] = D + dp[x][y][t]
            heapq.heappush( q,(dp[x][y][nexT] , x , y , nexT) )
print (-1)