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.
if __name__ == '__main__':
    input()

    arr = list(map(int, input().split()))

    m_arr = [(1, 1), (2 if arr[-2] < arr[-1] else 1, 1)]

    for n in range(len(arr) - 3, -1, -1):
        t0 = (m_arr[-1][0] + 1) if arr[n] < arr[n + 1] else 1
        t1 = (m_arr[-2][0] + 1) if arr[n] < arr[n + 2] else 1
        m_arr.append((t0, t1))

    idx = 0

    while idx < len(m_arr):
        elem = m_arr[idx]
        if elem[1] > elem[0]:
            m_arr[idx] = (elem[1], elem[1])
            while idx + 1 < len(m_arr) and m_arr[idx + 1][0] > 1:
                idx += 1
                m_arr[idx] = (m_arr[idx - 1][0] + 1, 0)
        idx += 1

    print(max(map(max, m_arr)))
