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.
# 20200114 19:38 ~ 20:00

n = int(input())
arr = list(map(int, input().split()))

incLen = list()
incS = list()

tempAns = 1

last = arr[0]
currS = 0
currLen = 1
for i in range(1, n):
    curr = arr[i]
    if curr > last:
        last = curr
        currLen += 1
    else:
        tempAns = max(tempAns, currLen)
        incLen.append(currLen)
        incS.append(currS)
        last = arr[i]
        currS = i
        currLen = 1
tempAns = max(tempAns, currLen)
incLen.append(currLen)
incS.append(currS)

for i in range(0, len(incLen)-1):
    prevLast = arr[incS[i] + incLen[i] - 1]
    nextFirst = arr[incS[i+1]]
    if incS[i] + incLen[i] - 2 >= 0 and incLen[i] >=2:
        prevLLast = arr[incS[i] + incLen[i] - 2]
        if nextFirst > prevLLast:
            tempAns = max(tempAns, incLen[i] - 1 + incLen[i+1])
    if incLen[i+1] >=2:
        nextFFirst = arr[incS[i+1]+1]
        if nextFFirst > prevLast:
            tempAns = max(tempAns, incLen[i] + incLen[i+1] - 1)

print(tempAns)