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 atexit, io, sys, collections, math, heapq
buffer = io.BytesIO() 
sys.stdout = buffer
@atexit.register 
def write(): sys.__stdout__.write(buffer.getvalue())   

##################################### 


n = raw_input()
ais = map(int, raw_input().split())


prev =[1 for _ in range(len(ais))]
suff =[1 for _ in range(len(ais))]
for i in range(1,len(ais)):
	if ais[i] > ais[i-1]:
		prev[i] += prev[i-1]
for i in range(len(ais)-2,-1,-1):
	if ais[i] < ais[i+1]:
		suff[i] += suff[i+1]
ans = max(prev)

for i in range(1, len(ais)-1):
	#delete i
	if ais[i-1] < ais[i+1]:
		ans = max(ans, prev[i-1] + suff[i+1])
print ans


