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() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<pair<int, int> > inc;
  for (int i = 0; i < n - 1; ++i) {
    int b = i;
    while (i < n - 1 && a[i + 1] > a[i]) ++i;
    if (b != i) inc.emplace_back(b, i);
  }
  int res = 1;
  for (auto i : inc) res = max(res, i.second - i.first + 1);
  for (int i = 0; i < inc.size() - 1 && !inc.empty(); ++i) {
    if (inc[i].second + 1 == inc[i + 1].first) {
      int d = inc[i].second;
      if ((d - 1 >= 0 && a[d - 1] < a[d + 1]) || (d + 2 < n && a[d] < a[d + 2]))
        res = max(res, inc[i + 1].second - inc[i].first);
    }
  }
  cout << res;
}
