You are given an array a consisting of n integers.

You can remove at most one element from this array. Thus, the final length of the array is n-1 or n.

Your task is to calculate the maximum possible length of the strictly increasing contiguous subarray of the remaining array.

Recall that the contiguous subarray a with indices from l to r is a[l ... r] = a_l, a_{l + 1}, ..., a_r. The subarray a[l ... r] is called strictly increasing if a_l < a_{l+1} < ... < a_r.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the maximum possible length of the strictly increasing contiguous subarray of the array a after removing at most one element.

Examples

Input


5
1 2 5 3 4


Output


4


Input


2
1 2


Output


2


Input


7
6 5 4 3 2 4 3


Output


2

Note

In the first example, you can delete a_3=5. Then the resulting array will be equal to [1, 2, 3, 4] and the length of its largest increasing subarray will be equal to 4.
from sys import stdin, stdout
cin = stdin.readline
cout = stdout.write

n = int(cin())
a = list(map(int, cin().split()))

length = i = 1
k = 0
x = [1]

while i < n:
	j = i
	while j < n:
		if a[j] > a[j-1]:
			j += 1
			x.append(j-i+1)
		else:
			x.append(1)
			break
	i = j + 1

x.append(1)
#print(x)
length = max(x)

lst = []
for i in range(1, n+1):
	if x[i] == 1:
		lst.append(i)

for i in range(len(lst)-1):
	if lst[i] >= n-1:
		break
	#print('i ', lst[i], i)
	if a[lst[i]+1] > a[lst[i]-1]:
		length = max(length, x[lst[i+1]-1]+x[lst[i]-1]-1)
		#print(i, length, 100)
	if lst[i] - 2 >= 0 and a[lst[i]] > a[lst[i]-2]: #and x[lst[i]+1] > 1
		#print(x[lst[i+1]-1], x[lst[i]-2])
		length = max(length, x[lst[i+1]-1]+x[lst[i]-2])
		#print(i, length, 200)

#print(length)
cout(str(length))



'''
while i < n:
	print('i ', i)
	j = i
	while j < n:
		if a[j] > a[j-1]:
			print(1000000, a[j], a[j-1], j)
			j += 1
			
		else:
			break
	print('x ', x)
	if k > 0 and a[i] > a[k]:
		length = max(length, j-i+x[-1], j-i+1)
	else:
		length = max(length, j-i+1)
	x.append(j-i+1)
	k = j - 1
	i =  k + 2

cout(str(length))
'''