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.
import sys

n=int(sys.stdin.readline())

a=[int(i) for i in sys.stdin.readline().split()]

if(n==1):
	print(1)
elif(n==2):
	if(a[0]<a[1]):
		print(2)
	else:
		print(1)
else:

	arr=[0]*n
	par=[-1]*n

	# do not remove any element
	ctr=1
	ans=1
	ind=0
	for g in range(n-1):
		if(a[g]<a[g+1]):
			ctr+=1
		else:
			arr[ind]=ctr
			par[ind]=ind
			ans=max(ans,ctr)
			ctr=1
			ind=g+1

	arr[ind]=ctr
	par[ind]=ind
	ans=max(ans,ctr)
	
	for h in range(n-1):
		if(arr[h+1]==0):
			arr[h+1]=arr[h]
		if(par[h+1]==-1):
			par[h+1]=par[h]

	# remove element
	for p in range(n-2):
		# if(a[p]>=a[p+1] and a[p]<a[p+2]):
			# ans=max(ans,arr[p]+arr[p+2])
		if(a[p]<a[p+2] and par[p]!=par[p+2]):
			val=arr[p]+arr[p+2]
			if(a[p]<a[p+1] or a[p+1]<a[p+2]):
				val-=1
			ans=max(ans,val)

	print(ans)
