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())

array = input()
array = [ int(x) for x in array.split() ]
L = len(array)

if n == 1:
    print(1)
    exit()
elif n == 0:
    print(0)
    exit()

A = [[1, -1], [1, -1]]
nj = 1
if array[1] > array[0]:
    nj += 1
A[1][0] = nj
ans = nj

for i in range(2, L):

    if array[i] > array[i-1]:
        nj += 1
    else:
        nj = 1

    A.append([nj, -1])

    if array[i] > array[i-2]:
        j = A[i-2][0] + 1
        jj = 1
        if array[i] > array[i-1]:
            jj = A[i-1][1] + 1
        A[i][1] = max([j, jj])
    else:
        if array[i] > array[i-1] and A[i-1][1] != -1:
            A[i][1] = A[i-1][1] + 1

    #print(A)

    m = max(A[i])
    if ans < m:
        ans = m

print(ans)
