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.
class X:

    def __init__(self, d):
        self.start = d
        self.end = d
        self.before_end = d
        self.after_start = d
        self.size = 1

    def add(self, d):
        self.before_end = self.end
        if self.after_start == self.start:
            self.after_start = d
        self.end = d
        self.size += 1


def compare(a, b):
    res = [a.size, b.size]
    if a.before_end < b.start or a.end < b.after_start:
        res.append(a.size + b.size - 1)

    return max(res)


if __name__ == "__main__":
    n = int(input())
    arr = [int(x) for x in input().split()]
    l = [X(arr[0])]
    l_index = 0

    # build contacted list
    for i in range(1,n):
        if arr[i] > l[l_index].end:
            l[l_index].add(arr[i])
        else:
            l_index += 1
            l.append(X(arr[i]))

    res = l[0].size
    for i in range(len(l) - 1):
        res = max(res, compare(l[i], l[i+1]))

    print(res)
