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()))+[0]

# left incl, right excl.
seq = []
cur_start = 0
prev = 0
for i, x in enumerate(a):
    if prev < x:
        pass
    else:
        seq.append((cur_start, i))
        cur_start = i
    prev = x

    
best = max([r-l for l, r in seq])
for (l1, r1), (l2, r2) in zip(seq, seq[1:]):
    if a[l2-1] < a[l2+1]:
        best = max(best, r2-l1-1)
    if a[l2-2] < a[l2]:
        best = max(best, r2-l1-1)
print(best)
