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())
li=list(map(int,input().split()))
arr=[]
p=0
b=[]
ans=[]
for i in range(n-1):
    if p==0:
        b.append(i)
        p=1
    if li[i]<li[i+1]:
        p+=1
    else:
        b.append(i)
        arr.append(b)
        ans.append(p)
        b=[]
        p=0
if p!=0:
    b.append(i+1)
    arr.append(b)
    ans.append(p)
else:
    b.append(i+1)
    b.append(i+1)
    arr.append(1)
ans1=max(ans)
for i in range(len(ans)-1):
    if arr[i+1][0]-arr[i][1]==1:
        if arr[i][1]!=0:
            if li[arr[i][1]-1]<li[arr[i+1][0]]:
                ans1=max(ans1,ans[i]-1+ans[i+1])
        if arr[i+1][0]!=n-1:
            if li[arr[i][1]]<li[arr[i+1][0]+1]:
                ans1=max(ans1,ans[i]-1+ans[i+1])
print(ans1)
            