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())
xs = [int(x) for x in input().split()]

# dp0[i] = best up to this point without any skips.
# dp1[i] = best up to this point having used a skip.
dp0 = [1 for _ in range(n)]
dp1 = [1 for _ in range(n)]

# Stop at n since there is no point removing the last element.
for i in range(1, n):
    # Previous + 1 if can add it, else reset
    dp0[i] = dp0[i - 1] + 1 if xs[i] > xs[i - 1] else 1

    if i > 1:
        # Continue having skipped previously
        left = dp1[i - 1] + 1 if xs[i] > xs[i - 1] else 1
        # Try skipping
        right = dp0[i - 2] + 1 if xs[i] > xs[i - 2] else 1
        dp1[i] = max(left, right)


# Could probably just best the max(dp1)
best = max(max(dp0), max(dp1))
print(best)
