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())
v = list(map(int,input().split()))
a = [0]*(n+1)
a[0] = 1;
for i in range(1,n):
	if v[i] <= v[i-1]:
		a[i] = 1
	else:
		a[i] = a[i-1]+1
ans = [1]
i = 1
while i < n-1:
	if a[i] == 1:
		if v[i-1] < v[i+1]:
			j = a[i-1]
			k = i+1
			while a[k] > 1:
				j = j+1
				i = k
				k = k+1
				if k == n:
					break
			ans.append(j)
		elif i > 1:
			if v[i-2] < v[i]:
				j = a[i-2]+1
				k = i
				if k < n-1:
					while v[k]<v[k+1]:
						j = j+1
						i = k
						k = k+1
						if k >=n-1:
							break
				ans.append(j)
	i = i+1
	if i >=n:
		break
f = [max(ans),max(a)]
print(max(f))
