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, math

def input():
    return sys.stdin.readline()[:-1]

def main():
    n = int(input())
    a = list(map(int,input().split()))
    dp = [[0, 0] for k in range(n+1)]

    dp[0][0] = 1
    for k in range(1,n):
        if a[k-1] < a[k]:
            dp[k][0] = dp[k-1][0] + 1
        else:
            dp[k][0] = 1

    for k in range(2,n):
        if a[k-1] < a[k]:
            dp[k][1] = dp[k-1][1] + 1
        if a[k-2] < a[k]:
            dp[k][1] = max(dp[k][1],dp[k-2][0]+1)

    ans = 0
    for e in dp:
        ans = max(ans,max(e))
    print(ans)

if __name__ == '__main__':
    main()
