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;
const int N = 2e5;
int df[N + 100], dr[N + 100], arr[N + 100];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", arr + i);
  }
  df[1] = 1;
  int ans = 1;
  for (int i = 2; i <= n; i++) {
    if (arr[i] > arr[i - 1]) {
      df[i] = df[i - 1] + 1;
    } else {
      df[i] = 1;
    }
    ans = max(ans, df[i]);
  }
  dr[n] = 1;
  for (int i = n - 1; i > 0; i--) {
    if (arr[i] < arr[i + 1]) {
      dr[i] = dr[i + 1] + 1;
    } else {
      dr[i] = 1;
    }
  }
  for (int i = 2; i < n; i++) {
    if (arr[i - 1] < arr[i + 1]) {
      ans = max(ans, df[i - 1] + dr[i + 1]);
    }
  }
  printf("%d\n", ans);
}
