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 sys
n=int(sys.stdin.readline().strip())
a=[int(i) for i in sys.stdin.readline().strip().split()]
#Process
a.append(-2e9)
dp=[0]*(n+1)
dp[0]=1
bg=0
for i in range(1,n+1):
	if a[i]>a[i-1]:
		dp[i]=dp[i-1]+1
	else:
		for j in range(bg,i):
			dp[j]=dp[i-1]
		dp[i]=1
		bg=i
cnt=0
for i in range(1,n):
	cnt=max(cnt,dp[i])
	if a[i-1]>=a[i] and a[i]<a[i+1] and a[i-1]<a[i+1]:
		cnt=max(cnt,dp[i-1]+dp[i+1]-1)
	if a[i-1]<=a[i] and a[i]>a[i+1] and a[i-1]<a[i+1]:
		cnt=max(cnt,dp[i-1]+dp[i+1]-1)
# print(dp)
sys.stdout.write(str(cnt))