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.
n=int(input())
a=list(map(int,input().split()))
diff1=[]
for i in range(1,n):
	diff1.append(a[i]-a[i-1])
from collections import deque
diff2=deque()
for i in range(n-2,-1,-1):
	diff2.appendleft(a[i]-a[i+1])

def aze(d):
	i=0
	j=1
	ans=[]
	while i<n-1:
		if d[i]<=0:
			ans.append(j)
			j=1
		else:
			ans.append(j)
			j+=1
			

		i+=1
	return(ans)

def aze2(d):
	i=n-2
	j=1
	ans=deque()
	while i>-1:
		if d[i]>=0:
			ans.appendleft(j)
			j=1
			
		else:
			ans.appendleft(j)
			j+=1
			
		i-=1
	return(ans)

ans1=aze(diff1)
ans2=list(aze2(list(diff2)))
m=max(ans2[0],ans1[n-2])
cnt=1
for i in range(1,n):
	if a[i]>a[i-1]:
		cnt+=1

	else:
		
		cnt=1
	m=max(m,cnt)
for i in range(1,n-1):
	if a[i-1]<a[i+1]:m=max(m,ans1[i-1]+ans2[i])
	else:m=max([m,ans1[i-1],ans2[i]])
print(m)