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

start = 0
prev = arr[0]
subarrays = []
for i, v in enumerate(arr):
    if i==0:
        continue
    if v <= prev:
        if i - start > 1:
            subarrays.append((start, i-1))
        start = i
    prev = v
i += 1
if i - start > 1:
    subarrays.append((start, i-1))
start = i


c_max = 1
for subarray in subarrays:
    c_max = max(c_max, subarray[1] - subarray[0] + 1)

for j in range(len(subarrays)-1):
    left = subarrays[j]
    right = subarrays[j+1]
    if right[0] == left[1]+1 and arr[right[0]+1] > arr[left[1]]:
        c_max = max(c_max, left[1]-left[0]+1 + right[1]-right[0])
    if right[0] == left[1]+1 and arr[right[0]] > arr[left[1]-1]:
        c_max = max(c_max, left[1]-left[0]+1 + right[1]-right[0])
print(c_max)
