You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.

The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).

The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).

Output

Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.

Examples

Input

5 4
1 3 5 7 9
6 4 2 8


Output

3 2 1 4


Input

5 5
1 2 1 2 5
3 1 4 1 5


Output

4 2 4 2 5
import java.util.Arrays;
import java.util.Scanner;

public class Main2 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int a = scanner.nextInt();
        int b = scanner.nextInt();
        int[] n = new int[a + 1];
        int[] m = new int[b];
        for (int i = 0; i < a; i++)
            n[i] = scanner.nextInt();
        n[a] = Integer.MAX_VALUE;
        for (int i = 0; i < b; i++)
            m[i] = scanner.nextInt();
        Arrays.parallelSort(n);
        int last = 0;
        for (int i = 0; i < b; i++) {
            if (i == 0) {
                last = binarySearch(n, 0, a, m[i]);
            } else {
                if (m[i] > m[i - 1]) {
                    last = binarySearch(n, last, a, m[i]);
                } else if (m[i] < m[i - 1]) {
                    last = binarySearch(n, 0, last, m[i]);
                }
            }
            System.out.print(last + " ");
        }
    }

    private static int binarySearch(int[] array, int low, int high, int key) {
        while (low < high) {
            int m = low + (high - low) / 2;
            if (array[m] <= key) {
                low = m + 1;
            } else if (array[m] > key) {
                high = m;
            }
        }
        return low;
    }
}
		    		 		   	 			  			 			 	