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.*;

public class A321{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] array = new int[N];

        for(int i = 0; i < N; i++){
            array[i] = sc.nextInt();
        }

        printLogestIncSubArr(array, array.length);
    }
    public static void printLogestIncSubArr(int arr[],
                                              int n)
    {
        // 'max' to store the length of longest
        // increasing subarray
        // 'len' to store the lengths of longest
        // increasing subarray at diiferent
        // instants of time
        int max = 1, len = 1, maxIndex = 0;

        // traverse the array from the 2nd element
        for (int i = 1; i < n; i++)
        {
            // if current element if greater than
            // previous element, then this element
            // helps in building up the previous
            // increasing subarray encountered
            // so far
            if (arr[i] >= arr[i-1])
                len++;
            else
            {
                // check if 'max' length is less
                // than the length of the current
                // increasing subarray. If true,
                // then update 'max'
                if (max < len)
                {
                    max = len;

                    // index assign the starting
                    // index of longest increasing
                    // contiguous subarray.
                    maxIndex = i - max;
                }

                // reset 'len' to 1 as from this
                // element again the length of the
                // new increasing subarray is
                // being calculated
                len = 1;
            }
        }

        // comparing the length of the last
        // increasing subarray with 'max'
        if (max < len)
        {
            max = len;
            maxIndex = n - max;
        }

        // Print the elements of longest
        // increasing contiguous subarray.
        System.out.println(max);
    }
}


