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
n=int(sys.stdin.readline())
arr=list(map(int,sys.stdin.readline().split()))
skip=[-1]*n
noskip=[-1]*n
noskip[0]=1
prevskip=[-1]*n
ans=1
z=True
for i in range(1,n):
    if arr[i]>arr[i-1]:
        noskip[i]=noskip[i-1]+1
        if z:
            skip[i]=skip[i-1]+1
        else:
            if arr[i]>arr[i-2]:
                skip[i]=skip[i-1]+1
            else:
                skip[i]=0
            z=True
        prevskip[i]=prevskip[i-1]+1
    else:
        if i==1:
            skip[i]=1
            noskip[i]=1
            prevskip[i]=1
        else:
            if arr[i]>arr[i-2]:
                #skip[i]=noskip[i-1]
                prevskip[i]=noskip[i-2]+1
            noskip[i]=1
            skip[i]=noskip[i-1]
            z=False
    ans=max(ans,skip[i],noskip[i],prevskip[i])
#print(skip,'skip')
#print(noskip,'noskip')
#print(prevskip,'prevskip')
print(ans)
