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.
import sys
input = sys.stdin.readline
N = int(input())
a = list(map(int, input().split()))

inf = N + 1
dp = [inf] * N
dp[0] = 0
ups = []
dws = []
for i in range(N):
  x = a[i]

  f = 1
  while len(ups) > 0 and a[ups[-1]] >= x:
    j = ups.pop()
    dp[i] = min(dp[i], dp[j] + 1)
    f = a[j] != x
  if f and len(ups) > 0: dp[i] = min(dp[i], dp[ups[-1]] + 1)

  f = 1
  while len(dws) > 0 and a[dws[-1]] <= x:
    j = dws.pop()
    dp[i] = min(dp[i], dp[j] + 1)
    f = a[j] != x
  if f and len(dws) > 0: dp[i] = min(dp[i], dp[dws[-1]] + 1)

  ups.append(i)
  dws.append(i)

print(dp[-1])