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 = input()

l = map(int, raw_input().split())

pre = [1]*n
post = [1]*n

for i in xrange(1,n):
    if l[i]>l[i-1]:
        pre[i] = pre[i-1]+1

for i in xrange(n-2,-1,-1):
    if l[i]<l[i+1]:
        post[i] = post[i+1]+1


maxi = 1

for i in xrange(1,n):
    front = 0
    if l[i-1]<l[i]:
        front = pre[i]-1
    back = 0
    if i+1<n and l[i]<l[i+1]:
        back = post[i]-1
    
    maxi = max(maxi, front+back+1)

    if i+1<n and l[i-1]<l[i+1]:
        maxi = max(maxi, pre[i-1]+post[i+1])

    maxi = max(maxi, pre[i])
    maxi = max(maxi, post[i])


print maxi

