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
input = stdin.readline
from heapq import heapify,heappush,heappop,heappushpop
from collections import defaultdict as dd, deque as dq,Counter as C
from math import factorial as f ,ceil,gcd,sqrt,log
from bisect import bisect_left as bl ,bisect_right as br
from itertools import combinations as c,permutations as p
from math import factorial as f ,ceil,gcd,sqrt,log
mp = lambda : map(int,input().split())
it = lambda: int(input())
ls = lambda : list(input().strip())
mt = lambda  r : [ list(mp()) for _ in range(r)]
lcm = lambda a,b : (a*b)//gcd(a,b)
def fibo_n(n):
	return (((1+sqrt(5))/2)**n)/sqrt(5)
mod = 1000000007
a = it()
b = list(mp())
v1 =[1]*a
mx =0
for i in range(1,a):
	if b[i]>b[i-1]:
		v1[i] = v1[i-1] +1
	mx = max(v1[i],mx)
v2 = [1]*a
for j in range(a-2,-1,-1):
	if b[j]<b[j+1]:
		v2[j]= v2[j+1]+1
	mx = max(v2[j],mx)
for k in range(2,a-1):
	if b[k-1]<b[k+1]:
		mx = max(mx,v1[k-1]+ v2[k+1])
print(mx)