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 dp[200005][3], a[200005];
int main(int argc, char** argv) {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] > a[i - 1])
      dp[i][0] = dp[i - 1][0] + 1, dp[i][1] = dp[i - 1][1] + 1;
    else
      dp[i][0] = 1, dp[i][1] = 1;
    if (i - 1 && a[i] > a[i - 2]) dp[i][1] = max(dp[i][1], dp[i - 2][0] + 1);
    ans = max(dp[i][0], max(dp[i][1], ans));
  }
  cout << ans;
  return 0;
}
