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

dp = [[0]*len(els) for _ in range(2)]
dp[0][0] = 1
dp[1][0] = 1

for i in range(1, len(els)):
    if els[i] > els[i-1]:
        dp[0][i] = 1 + dp[0][i-1]
    else:
        dp[0][i] = 1

for i in range(1, len(els)):
    if els[i] > els[i-1]:
        prvi = dp[1][i-1] + 1
    else:
        prvi = 1
    
    if i >= 2 and els[i] > els[i-2]:
        drugi = 1 + dp[0][i-2]
    else:
        drugi = 1

    dp[1][i] = max(prvi, drugi)

# for line in dp:
#     print(line)

print(max(max(dp)))
    