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())
a =list(map(int,input().split()))
ls = []
i = 0
while(i<len(a)):
    b = []
    b.append(a[i])
    i+=1
    while(i<len(a)):
        if(b[-1]<a[i]):
            b.append(a[i])
        else:
            break
        i+=1
    ls.append(b)
    # print(b)
# print(ls)
mx = len(ls[0])

for i in range(len(ls)-1):
    # print(i,i+1,ls[i],ls[i+1])
    val=1
    if(len(ls[i+1])>1):
        if(ls[i][-1]<ls[i+1][1]):
            val = len(ls[i]+ls[i+1])-1
    
    if (len(ls[i])>1):
        if(ls[i][-2]<ls[i+1][0]):
            val = len(ls[i]+ls[i+1])-1  
    val= max(val,len(ls[i]),len(ls[i+1]))
    if(val>mx):
        mx=val
print(mx)
        
