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 = list(map(int, input().split()))
subarray = []
cur_length = 1
for i in range(n - 1):
    if a[i] < a[i + 1]:
        cur_length += 1
    else:
        subarray.append(cur_length)
        cur_length = 1
subarray.append(cur_length)
result = max(subarray)
index_acc = 0
for i in range(len(subarray) - 1):
    index_acc += subarray[i]
    if subarray[i] >= 2 and subarray[i + 1] >= 2:
        leftmost = index_acc - 2
        rightmost = index_acc + 1
        if a[leftmost] < a[rightmost - 1] or a[leftmost + 1] < a[rightmost]:
            result = max(result, subarray[i] + subarray[i + 1] - 1)
print(result)
