There are n beautiful skyscrapers in New York, the height of the i-th one is h_i. Today some villains have set on fire first n - 1 of them, and now the only safety building is n-th skyscraper.

Let's call a jump from i-th skyscraper to j-th (i < j) discrete, if all skyscrapers between are strictly lower or higher than both of them. Formally, jump is discrete, if i < j and one of the following conditions satisfied: 

  * i + 1 = j 
  * max(h_{i + 1}, …, h_{j - 1}) < min(h_i, h_j) 
  * max(h_i, h_j) < min(h_{i + 1}, …, h_{j - 1}). 



At the moment, Vasya is staying on the first skyscraper and wants to live a little longer, so his goal is to reach n-th skyscraper with minimal count of discrete jumps. Help him with calcualting this number.

Input

The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — total amount of skyscrapers.

The second line contains n integers h_1, h_2, …, h_n (1 ≤ h_i ≤ 10^9) — heights of skyscrapers.

Output

Print single number k — minimal amount of discrete jumps. We can show that an answer always exists.

Examples

Input


5
1 3 1 4 5


Output


3

Input


4
4 2 2 4


Output


1

Input


2
1 1


Output


1

Input


5
100 1 100 1 100


Output


2

Note

In the first testcase, Vasya can jump in the following way: 1 → 2 → 4 → 5.

In the second and third testcases, we can reach last skyscraper in one jump.

Sequence of jumps in the fourth testcase: 1 → 3 → 5.

from collections import *
import sys  
# "". join(strings) 
#	print(guess, flush= True) 
def ri():
    return int(input())
 
def rl():
    return list(map(int, input().split()))

n = ri()
h = rl()

#graph: g[i] is the list of accessible points from building i
g = defaultdict(list)

#need to do it twice, once for h, then once for -h
#so we can take care of both conditions:
# max(hi+1,...,hj-1) < min(hi,hj) in the first passage
# max(-hi+1,...,-hj-1) < min(-hi,-hj)  in the second one , which is the same as : min(hi+1,...,hj-1) > max(hi,hj)
for rot in range(2):
	#we want to maintain a  monotonic stack ie:
	#h[st[0]] > h[st[1]] > ... > h[st[-1]]
	stack = []

	#we go left to right
	for i in range(n):
		while stack and h[stack[-1]] < h[i]:
			g[stack[-1]].append(i)
			stack.pop()

		#carefull with bulding of same height
		if stack:
			g[stack[-1]].append(i)
			if h[stack[-1]] == h[i]:
				stack.pop()

		stack.append(i)

	for i in range(n):
		h[i] = -h[i]


#BFS to find shortest path from 0 to any building i
dist = [-1]*n
dist[0] = 0
que = [0]
for b in que:
	for j in g[b]:
		if dist[j] == -1:
			que.append(j)
			dist[j] = dist[b] + 1

print(dist[n - 1])
