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())
mx = 1
a = input().split()
for i in range(n):
    a[i] = int(a[i])
pre = [0]*n
suf = [0]*n
pre[0] = 1
for i in range( 1, n):
    if(a[i] > a[i - 1]):
        pre[i] = pre[i - 1] + 1
    else:
        pre[i] = 1
    mx = max( mx, pre[i])
suf[n - 1] = 1
for i in range( n - 2, -1, -1):
    if(a[i] < a[i + 1]):
        suf[i] = suf[i + 1] + 1
    else:
        suf[i] = 1
for i in range(1 , n - 1):
    if(a[i - 1] < a[i + 1]):
        mx = max( mx, pre[i - 1] + suf[i + 1])
print(mx)   
	 		 				 	    	 	  	 		  				