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.
# Problem: D. Remove One Element
# Contest: Codeforces - Codeforces Round #605 (Div. 3)
# URL: https://codeforces.com/problemset/problem/1272/D
# Memory Limit: 256 MB
# Time Limit: 2000 ms
# 
# KAPOOR'S

from sys import stdin, stdout 

def INI():
	return int(stdin.readline())
	
def INL():
	return [int(_) for _ in stdin.readline().split()]
	
def INS():
	return stdin.readline()

def MOD():
    return pow(10,9)+7
	
def OPS(ans):
	stdout.write(str(ans)+"\n")
	
def OPL(ans):
	[stdout.write(str(_)+" ") for _ in ans]
	stdout.write("\n")

	
if __name__=="__main__":
	n=INI()
	X=INL()
	
	x=1
	D=[]
	for _ in range(n-1):
		if X[_]<X[_+1]:
			x+=1
		else:
			D.append([x,_])
			x=1
	
	D.append([x,_])
	d=len(D)
	ans=D[0][0]
	
	if n==2:
		if X[0]<X[1]:
			OPS(2)
		else:
			OPS(1)
	else:
		for _ in range(d-1):
			i=D[_][1]
			if i==0:
				if X[i]<X[i+2]:
					ans=max(ans,D[_][0]+D[_+1][0]-1)
				else:
					ans=max(ans,D[_][0])
			elif i==n-2 or i==n-1:
				ans=max(ans,D[_][0])
			else:
				if X[i]<X[i+2] or X[i-1]<X[i+1]:
					ans=max(ans,D[_][0]+D[_+1][0]-1)
				else:
					ans=max(ans,D[_][0])
		OPS(max(ans,D[-1][0]))