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 MAX_N = 2e5 + 10;
int n, ans = -1;
int a[MAX_N], dp[MAX_N][2];
int main() {
  cin >> n;
  for (int i = 2; i <= n + 1; i++) cin >> a[i];
  for (int i = 2; i <= n + 1; i++) {
    dp[i][0] = 1;
    if (a[i - 1] < a[i]) {
      dp[i][0] = max(dp[i][0], dp[i - 1][0] + 1);
      dp[i][1] = max(dp[i][1], dp[i - 1][1] + 1);
    }
    if (a[i - 2] < a[i]) {
      if (dp[i - 2][0]) {
        dp[i][1] = max(dp[i][1], dp[i - 2][0] + 1);
      }
    }
  }
  for (int i = 2; i <= n + 1; i++) ans = max(ans, max(dp[i][1], dp[i][0]));
  cout << ans << endl;
  return 0;
}
