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() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  map<int, int> mp;
  int prev = 0, count = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] > prev) {
      count++;
    } else {
      count = 1;
    }
    mp[i] = count;
    prev = arr[i];
  }
  int ans = 1;
  for (int i = 0; i < n; i++) {
    int j = i - mp[i] - 1;
    ans = max(mp[i], ans);
    if (j >= 0) {
      if (arr[j] < arr[j + 2]) ans = max(ans, mp[i] + mp[j]);
      int j = i - mp[i];
      if (arr[j] < arr[j + 2]) {
        ans = max(ans, mp[i] - 1 + mp[j]);
      }
    } else
      ans = max(ans, mp[i]);
  }
  cout << ans;
}
