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())
s = list(map(int,input().split(' ')))
dp = [[0]*2 for _ in range(N)]
ans = 1
dp[0][0] = 1;dp[0][1] = 1
dp[1][1] = 1;dp[1][0] = 1
if s[1]>s[0]:dp[1][0] = 2;ans = 2

for i in range(2,N):
    if s[i]>s[i-1] :
        dp[i][0] = dp[i-1][0]+1
        if s[i]>s[i-2]:
            dp[i][1] = max(dp[i-1][1], dp[i-2][0])+1
        else: dp[i][1] = dp[i-1][1]+1
    else:
        dp[i][0] = 1
        if s[i]>s[i-2]:
            dp[i][1] = dp[i-2][0]+1
        else:
            dp[i][1] = 1
    ans = max(ans,dp[i][0],dp[i][1])

print(ans)

