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.
q = int(input())
a = list(map(int, input().split()))
path = [1]
start = [0]
end = [0]
max_len = [1]
for i in range(1, len(a)):
    if a[i] > a[i - 1]:
        path.append(path[-1] + 1)
        max_len[-1] += 1
    else:
        path.append(1)
        start.append(i)
        max_len.append(1)

ans = max(max_len)
for i in range(len(start) - 1):
    if start[i] + max_len[i] - 2 >= 0 and max_len[i + 1] > 1:
        if a[start[i] + max_len[i] - 2] < a[start[i + 1]]:
            # print('FIRST')
            # print(a[start[i] + max_len[i] - 2], a[start[i + 1]])
            ans = max(ans, max_len[i] + max_len[i + 1] - 1)
        elif a[start[i] + max_len[i] - 1] < a[start[i + 1] + 1]:
            # print('SECOND')
            ans = max(ans, max_len[i] + max_len[i + 1] - 1)
            # print(a[start[i] + max_len[i] - 1], a[start[i + 1] + 1])
print(ans)