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.
import javax.print.DocFlavor;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.nio.file.Paths;
import java.util.*;
public class Main extends PrintWriter {
    static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));Main () { super(System.out); }public static void main(String[] args) throws IOException{ Main  d1=new Main   ();d1.main();d1.flush(); }

    void main() throws IOException {
        StringBuilder sb = new StringBuilder();
        int t=1;
     //  t=i(s()[0]);
        while(t-->0) {
            String[] s1=s();
            int n=i(s1[0]);int[] a=new int[n];arr(a,n);
            int[][] dp=new int[n][2];int ans=0;
            for(int i=0;i<n;i++){
                  if(i==0){
                      dp[i][0]=1;
                      dp[i][1]=0;
                      continue;
                  }
                   if(a[i]>a[i-1]){
                       dp[i][0]=dp[i-1][0]+1;
                       dp[i][1]=dp[i-1][1]+1;
                      if(i-2>=0&&a[i-2]<a[i]) dp[i][1]=Math.max(dp[i][1],Math.max(dp[i-1][0],dp[i-2][0]+1));
                      else dp[i][1]=Math.max(dp[i][1],dp[i-1][0]);
                  }else{
                       dp[i][0]=1;
                       if(i-2>=0&&a[i-2]<a[i])
                       dp[i][1]=dp[i-2][0]+1;
                   }
                   ans=Math.max(dp[i][0],Math.max(ans,dp[i][1]));
            }
            System.out.println(ans);
        }
        System.out.println(sb);
    }
    static String[] s() throws IOException { return s.readLine().trim().split("\\s+"); }
    static int i(String ss) {return Integer.parseInt(ss); }
    static long l(String ss) {return Long.parseLong(ss); }
    public void arr(int[] a,int n) throws IOException {String[] s2=s();for(int i=0;i<n;i++){ a[i]=i(s2[i]); }}
}
class Pair{
    int ind,val;
    public Pair(int a,int b){
        this.val=a;this.ind=b;
        }
}

