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__":
    n = int(input())
    arr = [int(x) for x in input().split()]
    if n is 2:
        print(2 if arr[1] > arr[0] else 1)
        exit()
    # l[i] is storing the length of strictly increasing sequence ending at index i
    l = [0] * (n + 1)
    # r[i] is storing the length of strictly increasing sequence starting at index i
    r = [0] * (n + 1)
    l[0] = 1
    r[n - 1] = 1
    for i in range(1, n):
        l[i] = l[i - 1] + 1 if arr[i] > arr[i - 1] else 1
    for i in range(n - 2, -1, -1):
        r[i] = r[i + 1] + 1 if arr[i] < arr[i + 1] else 1
        
    ans = max(max(l), max(r))
    for i in range(1, n - 1):
        if arr[i - 1] < arr[i + 1]:
            ans = max(ans, l[i - 1] + r[i + 1])
 
print(ans)