You've got an *n* × *m* pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.

A picture is a barcode if the following conditions are fulfilled: 
 -  All pixels in each column are of the same color. -  The width of each monochrome vertical line is at least *x* and at most *y* pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than *x* or greater than *y*.
The first line contains four space-separated integers *n*, *m*, *x* and *y* (1 ≤ *n*, *m*, *x*, *y* ≤ 1000; *x* ≤ *y*).

Then follow *n* lines, describing the original image. Each of these lines contains exactly *m* characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
from sys import stdin
input=lambda : stdin.readline().strip()
from math import ceil,sqrt,factorial,gcd
from collections import deque
n=int(input())
l=list(map(int,input().split()))
visited=set()
graph={i:set() for i in range(1,n+1)}
d={}
papa=[0 for i in range(n+1)]
level=[[] for i in range(n+1)]
z=[[0] for i in range(n+1)]
for i in range(n-1):
	a,b,c=map(int,input().split())
	graph[a].add(b)
	graph[b].add(a)
	d[(a,b)]=c
stack=deque()
# print(graph)
for i in graph:
	if len(graph[i])==1:
		stack.append([i,0])
m=0
while stack:
	# print(stack)
	x,y=stack.popleft()
	if len(graph[x])>=1:
		for i in graph[x]:
			t=i
			break
		if (t,x) in d:
			q=d[(t,x)]
		else:
			q=d[(x,t)]
		z[t].append(y+l[x-1]-q)
		graph[t].remove(x)
		if len(graph[t])==1:
			stack.append([t,max(z[t])])
for i in range(1,n+1):
	z[i].sort()
	if len(z[i])>=3:
		m=max(m,l[i-1]+z[i][-2]+z[i][-1])
	m=max(m,z[i][-1]+l[i-1])
print(m)