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 remove(n,a):
    i=0
    l=[]
    maxx=0
    while i<n:
        st=i
        end=i+1
        while end<n and a[end-1]<a[end]:
            end+=1
        l.append((st,end-1))
        diff=end-st
        if diff>maxx:
            maxx=diff
        i=end
    #print(l,maxx)
    i=1
    while i<len(l):
        summ=0
        if i!=len(l)-1 and l[i][0]==l[i][1]:
            if a[l[i+1][0]]>a[l[i-1][1]]:
                #print(i)
                summ=l[i-1][1]-l[i-1][0]+1 + l[i+1][1]-l[i+1][0]+1
            i+=1
        else:
            if a[l[i][0]]>a[l[i-1][1]-1] or (l[i][0]!=l[i][1] and a[l[i][0]+1]>a[l[i-1][1]]):
                #print(i)
                summ=l[i-1][1]-l[i-1][0]+l[i][1]-l[i][0]+1
        if summ>maxx:
            maxx=summ
        i+=1
    return maxx
n=int(input())
a=[int(i) for i in input().split()]
print(remove(n,a))