Problem: problem

N students from JOI High School are lined up in a row from east to west. The i-th student from the western end of the column is student i. Each student has a bib with one integer. Initially, the integer Ai is written on the number of student i.

There are M batons, and the batons are numbered from 1 to M. For k = 1, 2, ..., M, perform the following operations. The operation related to baton k (2 ≤ k ≤ M) is performed after the operation related to baton k -1 is completed.

1. The teacher gives the baton k to student 1.
2. Students who receive the baton pass the baton according to the following rules.
* Rule: Suppose student i receives the baton k.
* 1 ≤ i ≤ N-1: When the remainder of the integer of the number of student i divided by k is greater than the remainder of the integer of the number of student i + 1 divided by k, student i and student i + 1 exchanges the number, and student i passes the baton to student i + 1. Otherwise, student i passes the baton to student i + 1 without changing the bib.
* When i = N: Student N passes the baton to the teacher.
3. When the teacher receives the baton k from student N, the operation related to the baton k is finished.



Given the integer and the number of batons M that were originally written in the student number, write a program to find the integer of each student's number after the teacher receives the baton M from student N.

input

The input consists of 1 + N lines.

The integers N and M (1 ≤ N ≤ 100, 1 ≤ M ≤ 100) are written on the first line, separated by blanks, and represent the number of students and the number of batons, respectively.

The integer Ai (1 ≤ Ai ≤ 1000) is written on the i-th line (1 ≤ i ≤ N) of the following N lines, and represents the integer Ai first written on the number of student i.

output

The output consists of N lines. On line i (1 ≤ i ≤ N), output the integer of the number of student i after the teacher receives the baton M from student N.

Input / output example

Input example 1


6 4
3
2
8
3
1
Five


Output example 1


2
3
1
8
Five
3


Input example 2


10 6
1
2
3
Four
Five
6
7
8
9
Ten


Output example 2


6
1
2
3
Ten
Four
8
7
9
Five


Input / output example 1 has 6 students. Initially, the student numbers are 3, 2, 8, 3, 1, and 5, in that order. There are four batons.

* At the end of the operation related to baton 1, the student numbers are 3, 2, 8, 3, 1, and 5, respectively.
* At the end of the operation related to baton 2, the student numbers are 2, 8, 3, 3, 1, and 5, respectively.
* At the end of the operation related to baton 3, the student numbers are 2, 3, 3, 1, 8, and 5, respectively.
* At the end of the operation related to baton 4, the student numbers are 2, 3, 1, 8, 5, and 3 in order.



Creative Commons License

Information Olympics Japan Committee work "15th Japan Information Olympics JOI 2015/2016 Qualifying Competition Tasks"





Example

Input

6 4
3
2
8
3
1
5


Output

2
3
1
8
5
3

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;
import java.util.*;

public class Main {
	static final long MOD = 1000000007;
	public static void main(String[] args) {	
		FastScanner sc = new FastScanner();
		int N = sc.nextInt();
		int M = sc.nextInt();
		int[] A = new int[N];
		for(int i = 0; i < N; i++) {
			A[i] = sc.nextInt();
		}
		for(int i = 0; i < M; i++) {
			for(int j = 0; j < N; j++) {
				if(j == N-1) {
					continue;
				}
				else {
					if(A[j] % (i+1) > A[j+1] % (i+1)) {
						int tmp = A[j];
						A[j] = A[j+1];
						A[j+1] = tmp;
					}
				}
			}
		}
		for(int i = 0; i < N; i++) {
			System.out.println(A[i]);
		}
	}
	
}

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
    public boolean hasNext() { skipUnprintable(); return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt() {
    	return (int)nextLong();
    }
}