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.
import sys
ip=lambda :sys.stdin.readline().rstrip()
from collections import Counter
# for _ in range(int(ip())):
n=int(ip())
a=list(map(int,ip().split()))
dp=[[1,False] for _ in range(n)]
dp[0][0]=1
for i in range(1,n):
    if a[i]>a[i-1]:
        if dp[i][0]<1+dp[i-1][0]:
            dp[i][0]= 1+dp[i-1][0]
            if dp[i-1][1]:
                dp[i][1]=True
    if i-2>=0 and a[i]>a[i-2]:
        if not dp[i-2][1]:
            if dp[i][0]<1+dp[i-2][0]:
                dp[i][0]=1+dp[i-2][0]
                dp[i][1]=True
v=max(dp,key=lambda x:max(x[0],x[1]))
print(max(v))
