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 MOD = 1e9 + 7;
const int nax = 1e7 + 7;
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  vector<int> a(n, 1), b(n, 0), c(n, 1);
  int ans = 1;
  a[0] = 1;
  b[0] = 0;
  c[0] = 0;
  for (int i = 1; i < n; i++) {
    if (v[i] > v[i - 1]) {
      a[i] = a[i - 1] + 1;
      c[i] = max(c[i - 1], b[i - 1]) + 1;
    }
    if (v[i] > v[i - 2] or i == 1) {
      if (i == 1) {
        b[i] = 1;
      } else {
        b[i] = a[i - 2] + 1;
      }
    }
    ans = max({ans, a[i], b[i], c[i]});
  }
  cout << ans << endl;
  return 0;
}
