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.
from sys import stdin
# def rl():
#     return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations

# t = int(input())

n = int(input())
l = list(map(int,input().split()))
i = 0
maxi = 1
flag1 = 0
idx = []


for i in range(n-1):
    if l[i]>=l[i+1]:
        # print(l[i],idx[-1][0])
       try:
          idx.append([i,i - idx[-1][0]])
       except:
           idx.append([i,i+1])

cnt = 0
if idx == []:
    print(n)
    exit()
# print(idx)
for i in range(n-1):
    if l[i]>=l[i+1]:

        z = i+1
        a,b = idx[cnt]
        b-=1

        if cnt+1<len(idx):
          c = idx[cnt+1][1]

          if i-1>=0 and l[i-1]<l[i+1]:

             maxi = max(b+c,maxi)
          else:
              maxi = max(maxi,b+1,c)

          if i+2<n:
              b+=1
              c-=1

              if l[i+2]>l[i]:
                   maxi = max(b+c,maxi)

        else:
            c = n-1-i

            if i-1>=0 and l[i-1]<l[i+1]:

               maxi = max(b+c,maxi)
            else:

                maxi = max(maxi,b+1,c)

            if i+2<n:
                b+=1
                c-=1

                if l[i+2]>l[i]:
                    maxi = max(b+c,maxi)




        cnt+=1


print(maxi)

