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.Locale;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
	// write your code her
        Scanner sc = new Scanner(System.in).useLocale(Locale.ENGLISH);
        Scanner sc2 = new Scanner(System.in).useLocale(Locale.ENGLISH);
        String basura = sc.nextLine();
        String[] numbers = sc.nextLine().split("\\s+");
        long[] Numbers = new long[numbers.length];
        long[] Largos = new long[100001];
        long maximo = 1;
        long actual = 1;
        String temp;
        for(int i = 0; i < numbers.length; i++)
        {
            temp = numbers[i];
            Numbers[i] = Long.parseLong(temp);
        }
        for(int i = 0; i < Numbers.length - 1; i++)
        {
            if (Numbers[i] <= Numbers[i+1])
            {
                actual++;
            }
            else
            {
                actual = 1;
            }
            if (maximo < actual)
            {
                maximo = actual;

   
            }

        }

        System.out.println(maximo);
    }
}
