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())
l=list(map(int,input().split()))
ll=[]
dum=[0,l[0]]
i=1
mxx=-1
while i<n:
    if(l[i]>l[i-1]):
        dum.append(l[i])
    else:
        dum.append(i-1)
        ll.append(dum)
        mxx=max(mxx,len(dum)-2)
        dum=[i,l[i]]
    i+=1
if(len(dum)>=2):
    dum.append(n-1)
    ll.append(dum)
    mxx=max(mxx,len(dum)-2)
mx=-1
#print(ll)
for i in range(1,len(ll)):
    if(ll[i-1][-1]+1==ll[i][0] and len(ll[i])>3 and len(ll[i-1])>3):
        #print("lol")
        if(ll[i-1][-2]<ll[i][2]):
            if(len(ll[i-1])+len(ll[i])-4)>mx:
                mx=len(ll[i-1])+len(ll[i])-4
        elif(ll[i-1][-3]<ll[i][1]):
            #print("lol")
            if(len(ll[i-1])+len(ll[i])-4)>mx:
                mx=len(ll[i-1])+len(ll[i])-4
print(max(mxx,mx-1))       
'''
mx=l[0]
pmx=l[0]
i=1
mxx=-1
cou=1
lol=0
pcou=0
j=0
while i<n:
    if(l[i]>l[i-1]):
        mx=l[i]
        pmx=l[i-1]
        if(lol):
            pcou+=1
        cou+=1
    else:
        if(cou>mxx):
            mxx=cou
        if(pcou>mxx):
            mxx=pcou
        if(i+1<n and l[i+1]>pmx):
            lol=1
            pcou=cou+1
            cou=1
            mx=l[i]
            i+=1
        else:
            pcou=0
            cou=0
            lol=0
    i+=1
if(pcou>mxx):
    mxx=pcou
if(cou>mxx):
    mxx=cou
print(mxx)
'''