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 = [int(x) for x in input().split()]
B = [0] * n
k = 0
K = [1]
for i in range(1, n):
    if A[i-1] >= A[i]:
        k += 1
        K.append(0)
    K[k] += 1
    B[i] = k
m = max(K)

if n > 2:
    for i in range(1, n-1):
        mm = K[B[i-1]] + K[B[i+1]]
        if K[B[i]] == 1:
            if A[i-1] < A[i+1]:
                mm = K[B[i-1]] + K[B[i+1]]
                m = max(mm, m)
                # print (i,"mm", mm, A[i-1], A[i+1])
        else:
            if B[i-1]< B[i+1] and A[i-1] < A[i+1]:
                # print (i,"else", mm-1)
                m = max(mm - 1, m)
# print(A)
# print(B)
# print(K)
print(m)
