Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th day (1 ≤ i ≤ n) he makes ai money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence ai. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.

Help Kefa cope with this task!

Input

The first line contains integer n (1 ≤ n ≤ 105).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the length of the maximum non-decreasing subsegment of sequence a.

Examples

Input

6
2 2 1 3 4 1


Output

3

Input

3
2 2 9


Output

3

Note

In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.

In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
 
 
public class JavaApplication1 {
 
    public static void main(String[] args) {
        Scanner sc =new Scanner(System.in);
        int n =sc.nextInt();
        
        List<Long> list = new ArrayList<>();
         List<Integer> list1 = new ArrayList<>();
        int count = 0 ;
        
        for (int i = 0; i < n; i++) {
            long k =sc.nextLong();
            list.add(k);
            
            
        }
        for (int i = 0; i < n; i++) {
            if (i==0) {
                count = 1 ;
                 if (i == n-1) {
                    list1.add(count);
                }
            }else if (list.get(i) >= list.get(i-1)) {
                count++;
                 if (i == n-1) {
                    list1.add(count);
                }
            }else
            { list1.add(count);
           
              count = 1;
                if (i == n-1) {
                    list1.add(count);
                }
            
            }
         
        }
        
        Collections.sort(list1);
        if (list1.size()==0) {
            System.out.println(String.valueOf(0));
        }else
        {
             System.out.println(String.valueOf(list1.get(list1.size()-1)));
        }
        
        
   

 
       
        
        
        
    }
}