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.io.*; 
import java.util.Scanner;
 
public class KefaFirstSteps {
    public static Scanner in;
    public static void main(String[] args) throws FileNotFoundException {
        // PrintStream o = new PrintStream(new File("A.txt"));System.setOut(o);
        in = new Scanner(System.in);
        solve();
    }
    
    private static void solve() {
        int n = ni();
        int total = 0;
        int money = 0;
        int max = 0;
        for(int i=0;i<n; i++){
            int temp = ni();
            if(temp >= money){
                money  =temp;
                total++;
            }
            else{
                money = temp;
                max = Math.max(max, total);
                total = 1;
            }
        }
        // money = temp;
        max = Math.max(max, total);
        // total = 1;
        System.out.println(max);
    }
    
    private static int ni() {
        return in.nextInt();
    }
    
    private static int[] na(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) a[i] = ni();
        return a;
    }
    
    private static long[] nal(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) a[i] = nl();
        return a;
    }
    
    private static long nl() {
        return in.nextLong();
    }
    
    private float nf() {
        return in.nextFloat();
    }
    
    private static double nd() {
        return in.nextDouble();
    }
}