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 maxn = 2e5 + 5;
const int INF = 1e9;
int dat[maxn];
int dp[maxn];
int dp1[maxn];
void solve() {
  memset(dp, 0, sizeof(dp));
  int n;
  scanf("%d", &n);
  dat[0] = -1;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", dat + i);
    if (dat[i] > dat[i - 1]) {
      dp[i] = dp[i - 1] + 1;
    } else {
      dp[i] = 1;
    }
    ans = max(ans, dp[i]);
  }
  dp[n + 1] = 0;
  dat[n + 1] = -1;
  int up = n;
  int t = n;
  while (t > 0) {
    int tt = dp[t];
    for (int i = 1; i <= tt; i++) {
      dp1[t] = i;
      t--;
    }
  }
  for (int i = n; i >= 1; i--) {
    if (dat[i - 1] < dat[i + 1]) {
      ans = max(ans, dp[i - 1] + dp1[i + 1]);
    }
  }
  printf("%d\n", ans);
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
