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())
arr = [int(i) for i in input().split()]


def maxsub(a):
  inc = [[a[0]]]
  for x in a[1:]:
    if(x > inc[-1][-1]): inc[-1].append(x)
    else: inc.append([x])
  
  l = len(inc)
  for i in range(l-1):
    if len(inc[i+1])==1:
      if i+2<l:
        if inc[i][-1] < inc[i+2][0]:
          inc[i] = inc[i] + inc[i+2]
    else:
      can, can2 = inc[i], inc[i]
      if inc[i][-1] < inc[i+1][1]:
        can = inc[i] + inc[i+1][1:]
      if len(inc[i]) > 1:
        if inc[i][-2] < inc[i+1][0]:
          can2 = inc[i][:-1] + inc[i+1]

      if(len(can) > len(can2)): inc[i] = can
      else: inc[i] = can2
      
  
  for i in range(len(inc)):
    inc[i] = len(inc[i])
  
  return max(inc)

print(maxsub(arr))