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.
#設定
import sys
input = sys.stdin.buffer.readline

#ライブラリインポート
from collections import defaultdict

#入力受け取り
def getlist():
	return list(map(int, input().split()))

#処理内容
def main():
	N = int(input())
	A = getlist()

	DP1 = [1] * N
	DP2 = [1] * N
	for i in range(N - 1):
		if A[i + 1] > A[i]:
			DP1[i + 1] = DP1[i] + 1
	# print(DP1)

	for i in range(2, N - 1):
		if A[i] > A[i - 2]:
			DP2[i] = max(DP2[i], DP1[i - 2] + 1)
		if A[i + 1] > A[i]:
			DP2[i + 1] = DP2[i] + 1

	if N >= 3:
		if A[N - 1] > A[N - 3]:
			DP2[N - 1] = max(DP2[N - 1], DP1[N - 3] + 1)
	# print(DP2)

	ans = 0
	for i in range(N):
		ans = max(ans, DP1[i])
	for i in range(N):
		ans = max(ans, DP2[i])
	print(ans)

if __name__ == '__main__':
	main()