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 d4x[4] = {1, 0, -1, 0};
int d4y[4] = {0, 1, 0, -1};
int d8x[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int d8y[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long oo = 1e18;
const long long maxN = 1e6;
void maximize(int &a, int b) { a = max(a, b); }
void minimize(int &a, int b) { a = min(a, b); }
int a[maxN], L[maxN], R[maxN];
void solve() {
  int n, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) L[i] = R[i] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] > a[i - 1]) L[i] = L[i - 1] + 1;
    ans = max(ans, L[i]);
  }
  for (int i = n - 1; i >= 1; i--) {
    if (a[i] < a[i + 1]) R[i] = R[i + 1] + 1;
    ans = max(ans, R[i]);
  }
  for (int i = 2; i < n; i++) {
    if (a[i - 1] < a[i + 1]) {
      ans = max(ans, L[i - 1] + R[i + 1]);
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}
