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.Arrays;
import java.util.Scanner;

/**
 *
 * @author macbookpro
 */
public class A580 {
    public static void main(String[] args) {
        
        
                Scanner in = new Scanner(System.in);
                
                int n = in.nextInt();
                 
                int a[]= new int[n];
                int dp[]= new int[n];
                for(int i=0; i<n; i++)
                {
                  a[i]=in.nextInt();
                }
                Arrays.fill(dp, 1);
             
                for(int i=1; i<n;i++)
                {
                   if(a[i-1]<=a[i])
                   {
                       dp[i]+=dp[i-1];
                       
                       
                   }
                   else{
                       dp[i]=1;
                   }
                }
                int max=dp[0];
               for(int i=1; i<n;i++
                    )
               {
                if(max<dp[i])
                {
                    max=dp[i];
                }
                   
               }
                
                System.out.println(max);
                
                
               
                
                
    }
}
