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.
#Remove one element
n = int(input())
l = list(map(int,input().split()))
p = [0]*n
z = [0]*n
i = 0
d = 1
while i < len(l):
    cnt = 1
    pi = i
    pj = i+1
    while pj < len(l):
        if l[pj] > l[pi]:
            cnt += 1

        else:
            break

        pj += 1
        pi += 1

    j = i
    k1 = 1
    k2 = cnt
    while k1 <= cnt:
        p[j] = (k1,k2)
        z[j] = d
        k1 += 1
        k2 -= 1
        j += 1

    i = j
    d += 1

m = 0
#print(z)
#print(p)
for i in range(n):
    if i < 2:
        x = p[i][1]

    else:
        e1 = p[i][1]
        e2 = -1
        if l[i] > l[i-2] and z[i-2] != z[i]:
            e2 = e1+p[i-2][0]

        x = max(e1,e2)

    if x > m:
        m = x

print(m)