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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int l[n], r[n];
  for (long long int i = 0; i < n; i++) {
    l[i] = 1;
    r[i] = 1;
  }
  for (long long int i = 1; i < n; i++) {
    if (a[i] > a[i - 1]) r[i] = r[i - 1] + 1;
  }
  for (long long int j = n - 2; j >= 0; j--) {
    if (a[j] < a[j + 1]) l[j] = l[j + 1] + 1;
  }
  for (long long int j = 1; j <= n - 2; j++) {
    if (a[j - 1] < a[j + 1]) {
      l[j] = l[j + 1] + r[j - 1];
    }
  }
  cout << *max_element(l, l + n) << endl;
  return 0;
}
