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.*;
import java.util.LinkedList;

public class Solve {

    public static void swap(int a,int b){
        int temp = a;
        a=b;
        b=temp;
    }

    public static int gcd(int a, int b) {
       if (a == 0)
         return b;
       return gcd(b % a, a);
    }

    public static int lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }

    public static int sumOfDigits(int n){
        int sum=0;
        while (n!=0){
            sum+=n%10;
            n/=10;
        }
        return sum;
    }

    public static int ceil(int x,int y){
        return (x+y-1)/2;
    }
    public static int numberOfDigits(int n){
        int count=0;
        while(n!=0){
            count++;
            n/=10;
        }
        return count;
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int[] a = new int[n];
        for(int i=0;i<n;i++){
            a[i] = scan.nextInt();
        }
        int count = 1,max=1;
        int prev = a[0];
        for(int i=1;i<n;i++){
            if(prev<=a[i]) {
                count++;
                max = Math.max(max,count);
            }
            else{
                count=1;
            }
            prev = a[i];
        }
        System.out.println(max);
    }
}