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(i) for i in input().split()]

d = []

actual = [a[0]]

for i in range(1,n):
    if a[i] > a[i-1]:
        actual.append(a[i])
    else:
        d.append([len(actual), actual])
        actual = [a[i]]

d.append([len(actual), actual])

maxi = 0

if len(d) == 1:
    print(n)
else:
    n = len(d)
    for i in range(n):
        maxi = max(maxi,d[i][0])
    for i in range(n-1):
        if d[i][0] > 1 and d[i + 1][0] > 1:
            if d[i][1][-2] < d[i+1][1][0] or d[i][1][-1] < d[i+1][1][1]:
                maxi = max(maxi, d[i][0] + d[i+1][0] - 1)
    for i in range(n-2):
        if d[i+1][0] == 1:
            if d[i][1][-1] < d[i+2][1][0]:
                maxi = max(maxi,d[i][0] + d[i+2][0])
    print(maxi)