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())
arr = [int(x) for x in input().split()]
asc = [1]
for a1, a2 in zip(arr, arr[1:]):
    if a1 < a2:
        asc.append(asc[-1] + 1)
    else:
        asc.append(1)
des = [1]
for a1, a2 in zip(arr[:-1][::-1], arr[1:][::-1]):
    if a1 < a2:
        des.append(des[-1] + 1)
    else:
        des.append(1)
des = des[::-1]
max_len = [1] * n
for i in range(n):
    if i == 0:
        max_len[i] = asc[i+1]
    elif i == n-1:
        max_len[i] = des[i-1]
    elif arr[i+1] > arr[i - 1]:
        max_len[i] = asc[i-1] + des[i+1]
print(max(max_len + asc))
