Polycarp has to solve exactly n problems to improve his programming skill before an important programming competition. But this competition will be held very soon, most precisely, it will start in k days. It means that Polycarp has exactly k days for training!

Polycarp doesn't want to procrastinate, so he wants to solve at least one problem during each of k days. He also doesn't want to overwork, so if he solves x problems during some day, he should solve no more than 2x problems during the next day. And, at last, he wants to improve his skill, so if he solves x problems during some day, he should solve at least x+1 problem during the next day.

More formally: let [a_1, a_2, ..., a_k] be the array of numbers of problems solved by Polycarp. The i-th element of this array is the number of problems Polycarp solves during the i-th day of his training. Then the following conditions must be satisfied: 

  * sum of all a_i for i from 1 to k should be n; 
  * a_i should be greater than zero for each i from 1 to k; 
  * the condition a_i < a_{i + 1} ≤ 2 a_i should be satisfied for each i from 1 to k-1. 



Your problem is to find any array a of length k satisfying the conditions above or say that it is impossible to do it.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 10^5) — the number of problems Polycarp wants to solve and the number of days Polycarp wants to train.

Output

If it is impossible to find any array a of length k satisfying Polycarp's rules of training, print "NO" in the first line.

Otherwise print "YES" in the first line, then print k integers a_1, a_2, ..., a_k in the second line, where a_i should be the number of problems Polycarp should solve during the i-th day. If there are multiple answers, you can print any.

Examples

Input


26 6


Output


YES
1 2 4 5 6 8 


Input


8 3


Output


NO


Input


1 1


Output


YES
1 


Input


9 4


Output


NO
import java.util.*;

public class npkds {
    public static void main( String[] args ) {
	Scanner in = new Scanner( System.in );
	int n = in.nextInt(), k = in.nextInt();
	int mid;
	if ( k % 2 == 1 )
	    mid = n / k;
	else
	    mid = ( n + ( k / 2 ) ) / k;
	if ( mid <= k / 2 ) {
	    System.out.println( "NO" );
	    return;
	}
	mid -= k / 2;
	int[] a = new int[k];
	for ( int i = 0; i < k; i++ ) {
	    a[i] = mid + i;
	    n -= a[i];
	}
	int c = 1;
	if ( a[0] == 1 )
	    c = 2;
	while ( c != k && n != 0 ) {
	    for ( int i = k - 1; n != 0 && i >= c; i-- ) {
		a[i]++;
		n--;
	    }
	    if ( a[c] == 2 * a[c - 1] )
		c++;
	}
	if ( n != 0 )
	    System.out.println( "NO" );
	else {
	    System.out.println( "YES" );
	    for ( int i = 0; i < k; i++ )
		System.out.print( a[i] + " " );
	}
    }
}
