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())
arr = list(map(int, input().split(' ')))

counts = set()
last = -1

for n in range(N):
	count = 1
	delite = False
	delited = -1
	i = n
	j = i+1
	
	if n < last:
		continue
	
	while True:
		if j < N:
			if arr[i] < arr[j]:
				count += 1
				if not delite:
					last = i-1
			else:
				if not delite:
					if arr[i-1] < arr[j] and i > 1:
						delited = i
						count -= 1
						i -= 2
						j -= 1
						delite = True
					elif j+1 < N:
						if arr[i] < arr[j+1]:
							delited = j
							i -= 1
							delite = True
						else:
							counts.add(count)
							break
					else:
						counts.add(count)
						break
				else:
					counts.add(count)
					break
			i += 1
			j += 1
			if i == delited:
				i += 1
			if j == delited:
				j += 1
		else:
			counts.add(count)
			break

print(max(counts))