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.
# HEY STALKER
n = int(input())
l = list(map(int, input().split()))
ans = []
v = []
x = []
for t in range(len(l)):
    if len(v) == 0:
        v.append(l[t])
    elif l[t] > l[t-1]:
        v.append(l[t])
    else:
        ans.append(v)
        v = [l[t]]
    if t == len(l)-1:
        ans.append(v)
# print(*ans)
x.append(len(ans[-1]))
for i in range(len(ans)-1):
    x.append(len(ans[i]))
    if len(ans[i]) > 1 and ans[i][-2] < ans[i+1][0]:
        x.append(len(ans[i]) + len(ans[i+1]) - 1)
    if len(ans[i+1]) > 1 and ans[i+1][1] > ans[i][-1]:
        x.append(len(ans[i]) + len(ans[i+1]) - 1)
print(max(x))