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;
long long int powMod(long long int x, long long int y) {
  long long int p = 1;
  while (y) {
    if (y % 2) {
      p = (p * x) % 1000000007;
    }
    y /= 2;
    x = (x * x) % 1000000007;
  }
  return p;
}
long long int invMod(long long int x) { return powMod(x, 1000000007 - 2); }
void solve() {
  long long int n;
  cin >> n;
  long long int a[n], s[n], e[n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < n; i++) s[i] = 1, e[i] = 1;
  long long int Max = 0;
  for (long long int i = n - 2; i >= 0; i--) {
    if (a[i] < a[i + 1]) s[i] = s[i + 1] + 1;
    Max = max(Max, s[i]);
  }
  for (long long int i = 0; i < n - 1; i++) {
    if (a[i] < a[i + 1]) e[i + 1] = e[i] + 1;
    Max = max(Max, e[i + 1]);
  }
  for (long long int i = 0; i < n - 2; i++) {
    if (a[i] < a[i + 2]) Max = max(Max, e[i] + s[i + 2]);
  }
  cout << Max << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
