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.
#!/usr/bin/env python3
import sys, math, itertools, heapq, collections, bisect, string
input = lambda: sys.stdin.buffer.readline().rstrip().decode('utf-8')
inf = float('inf')
mod = 10**9+7
ans = 0 ;count = 0 ;pro = 1



n=int(input())
A=list(map(int,input().split()))
L=[0]*n
R=[0]*n
for i in range(n-1):
    count+=1
    L[i]=count
    if A[i]>=A[i+1]:
        count=0
L[n-1]=count+1
count=0
for i in range(n-1,0,-1):
    count+=1
    R[i]=count
    if A[i-1]>=A[i]:
        count=0
R[0]=count+1
# print(L)
# print(R)

if L[n-1]==R[0]==n:
    print(n)
    exit()

for i in range(1,n-1):
    if A[i-1]<A[i+1]:
        ans=max(ans,L[i-1]+R[i+1])
    else:
        ans=max(ans,L[i-1],R[i+1])
ans=max(ans,R[1])
ans=max(ans,L[n-2])
print(ans)
    