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(_) for _ in input().split()]
l1 = []
l2 = [a[0]]
l3 = []
for i in range(1, n):
    if a[i] > a[i - 1]:
        l2.append(a[i])
    else:
        l1.append(l2[:])
        l2.clear()
        l2.append(a[i])
l1.append(l2)
for i in l1:
    l3.append(len(i))
max_ = max(l3)
for i in range(1, len(l3) - 1):
    if l3[i] == 1 and l1[i - 1][-1] < l1[i + 1][0] and l3[i - 1] + l3[i + 1] > max_:
        max_ = l3[i - 1] + l3[i + 1]
for i in range(len(l3) - 1):
    shit = 0
    if l3[i] > 1 and l3[i + 1] > 1:
        if l1[i][-2] < l1[i + 1][0]:
            shit = l3[i] + l3[i + 1] - 1
        elif l1[i][-1] < l1[i + 1][1]:
            shit = l3[i] + l3[i + 1] - 1
    if shit > max_:
        max_ = shit
print(max_)
