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.
import sys
input = sys.stdin.readline


def main():
    N = int(input())
    A = [int(x) for x in input().split()]

    AX = []
    for i in range(N):
        if i == 0:
            AX.append(1)
        else:
            if A[i] > A[i - 1]:
                AX.append(AX[i - 1] + 1)
            else:
                AX.append(1)
    BX = [1] * N
    for i in range(N - 1, -1, -1):
        if i == N - 1:
            BX[i] = 1
        else:
            if A[i + 1] > A[i]:
                BX[i] = BX[i + 1] + 1
            else:
                BX[i] = 1

    ans = max(AX)
    for i in range(N):
        if i == 0 or i == N - 1:
            continue
        if A[i - 1] < A[i + 1]:
            ans = max(ans, AX[i - 1] + BX[i + 1])

    print(ans)






    
    

if __name__ == '__main__':
    main()


