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 ()))
f = [[1, 0] for i in range (200002)]
use = [False for i in range (200002)]
ans = 1

if (a[1] > a[0]) :
    f[1][0] = 2
    ans = 2

for i in range (2, len (a)) :
    u1 = 0
    u2 = 0
    v1 = 0

    if (a[i] > a[i - 1]) :
        u1 = f[i - 1][0]
        u2 = f[i - 1][1]
    if (a[i] > a[i - 2]) :
        v1 = f[i - 2][0]
    f[i][0] += u1
    f[i][1] = 1 + max (u2, v1)
    if (f[i][1] == 1) :
        f[i][1] = 0
    ans = max (f[i][0], f[i][1], ans)
    
print (ans)
    
 	  			 		 	 		 						 		   	 	