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.
# lET's tRy ThIS...

import math
import os
import sys

#-------------------BOLT------------------#
#-------Genius----Billionare----Playboy----Philanthropist----NOT ME:D----#

input = lambda: sys.stdin.readline().strip("\r\n")
def cin(): return sys.stdin.readline().strip("\r\n")
def fora(): return list(map(int, sys.stdin.readline().strip().split()))
def string(): return sys.stdin.readline().strip()
def cout(ans): sys.stdout.write(str(ans))
def endl(): sys.stdout.write(str("\n"))
def ende(): sys.stdout.write(str(" "))

#---------ND-I-AM-IRON-MAN------------------#

def right(n, a, i, cnt):
    j=i+2
    cnt+=1
    while(j<n-1):
        if(a[j+1]>a[j]):
            cnt+=1
        else:
            return cnt
        j+=1
    return cnt

def curr(n, a, i, cnt):
    j=i+1
    cnt+=1
    while(j<n-1):
        if(a[j+1]>a[j]):
            cnt+=1
        else:
            return cnt
        j+=1
    return cnt

def main():
    #for _ in range(int(input())):
    #LET's sPill the BEANS
    n=int(cin())
    a=fora()
    i=0 
    cnt=1 
    flag=1
    l=[]
    l.append(1)
    while(i<n-1):
        if(a[i+1]>a[i]):
            cnt+=1
        else:
            if(i<n-2 and a[i]<a[i+2]):
                    prevcnt=cnt
                    l.append(right(n,a,i,prevcnt))  
            if(cnt>1 and a[i-1]<a[i+1]):
                    prevcnt=cnt-1
                    l.append(curr(n,a,i,prevcnt))
            l.append(cnt)
            cnt=1 
        i+=1
    l.append(cnt)
    # print(l)
    # ende()
    cout(max(l))    
    endl()

if __name__ == "__main__": 
    main()
