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())
lis=[-1]+list(map(int,input().split()))+[0]
ans=[0]+[1]*(n)+[1]
prev=0
for i in range(1,n+1):
    if lis[i-1]<lis[i]:
        ans[i]+=ans[i-1]
k=max(ans)
prev=0
d=0
ind=0
for i in range(1,n+2):
    if ans[i]==1:
#        print(ind,prev,lis[ind+2],lis[ind],ans[i-1])
        if lis[ind+2]>lis[ind]:
            d=max(d,ans[i-1]+prev)
        if lis[ind+1]>lis[ind-1]:
            d=max(d,prev+ans[i-1])         
#        print(d,'d')    
        prev=ans[i-1]
        ind=i-1
d-=1        
print(max(k,d))        





