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())
maxx=1
ch=input()
nb=1
nb2=1
L=[int(i)for i in ch.split()]
L2=[]
L3=[]
test=False
for i in range(n-2):
          if L[i]<L[i+1]:
                    nb+=1
 
                    
          else:
                    L2.append(nb)
                    nb=1
                    if (L[i-1]>=L[i+1])and(L[i]>=L[i+2]):
                              L3.append(False)     
                              
                              
                    else:
                              L3.append(True)
                              
 
if L[n-2]<L[n-1]:
    nb+=1
 
L2.append(nb)
 
L4=[]
if len(L3)>0:
    for i in range(len(L3)):
        if L3[i]==True:
            L4.append(L2[i]+L2[i+1]-1)
        else:
            L4.append(L2[i])
            L4.append(L2[i+1])
    print(max(L4))
else:
    print(L2[0])