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.
def solve(arr,n):
    if(n==1):
        return 1
    if(n==2):
        if(arr[1]>arr[0]):
            return 2
        else:
            return 1
    dp=list()
    org=list()
    curr=1
    org.append(1)
    for i in range(1,n):
        if(arr[i]>arr[i-1]):
            curr+=1
            org.append(curr)
        else:
            curr=1
            org.append(curr)
    dp.append(org[0])
    dp.append(org[1])
    for i in range(2,n):
        if(arr[i]>arr[i-2]):
            if(arr[i]>arr[i-1]):
                dp.append(max(dp[i-1]+1,org[i-2]+1))
            else:
                dp.append(max(org[i],org[i-2]+1))
            
        else:
            if(arr[i]>arr[i-1]):
                dp.append(dp[i-1]+1)
            else:
                dp.append(org[i])
    #print(dp)
    return max(dp)


n=int(input())
arr=list(map(int,input().split()))
print(solve(arr,n))