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())
a=[int(x) for x in input().split()]
l=[1 for i in range(n)]
r=[1 for i in range(n)]
for i in range(1,n):
    if a[i]>a[i-1]:
        l[i]=l[i-1]+1
for i in range(n-2,0,-1):
    
    if a[i]<a[i+1]:
        r[i]=r[i+1]+1
maxi=-9999999999

p=max(max(l),max(r))
for i in range(n):
    if i==0:
        maxi=max(maxi,r[i+1])
    elif i==n-1:
        maxi=max(maxi,l[i-1])
    else:
        if a[i-1]>=a[i+1]:
            maxi=max(maxi,max(l[i-1],r[i+1]))
        else:    
            maxi=max(maxi,l[i-1]+r[i+1])
           
print(max(p,maxi))   