803_C. Maximal GCD_13669
You are given positive integer number n. You should create such strictly increasing sequence of k positive numbers a1, a2, ..., ak, that their sum is equal to n and greatest common divisor is maximal.

Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.

If there is no possible sequence then output -1.

Input

The first line consists of two numbers n and k (1 ≤ n, k ≤ 1010).

Output

If the answer exists then output k numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them.

Examples

Input

6 3


Output

1 2 3


Input

8 2


Output

2 6


Input

5 3


Output

-1
import java.io.*;
import java.util.StringTokenizer;

public class MaximalGCD {

	public static FastScanner in = new FastScanner(System.in);

	public static class FastScanner {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public FastScanner(InputStream i) {
			reader = new BufferedReader(new InputStreamReader(i));
			tokenizer = new StringTokenizer("");
		}

		public String next() throws IOException {
			while (!tokenizer.hasMoreTokens())
				tokenizer = new StringTokenizer(reader.readLine());
			return tokenizer.nextToken();
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public void close() throws IOException {
			reader.close();
		}
	}

	// variables

	public static long n, k;

	// onReads

	public static void main(String[] args) {
		try {
			n = in.nextLong();
			k = in.nextLong();
			// readValues
			// useFunctions
			// printResult
			long sumK = sum(k);
			if (k >= 141421 || n < sumK)
				System.out.println(-1);
			else {
				long ratio = n / sumK;
				if (n % ratio != 0)
					ratio = getNewRatio(n, ratio);
				for (int i = 1; i <= k - 1; i++) {
					System.out.print(ratio * i + " ");
					n -= ratio * i;
				}
				System.out.println(n);
			}
			in.close();
		} catch (Exception e) {
			e.printStackTrace();
		}
	}

	// functions

	public static long sum(long k) {
		return k * (k + 1) / 2;
	}

	public static long getNewRatio(long n, long maxRatio) {
		long result = 1;
		for (Long i = 1l; i * i <= n; i++)
			if (n % i == 0) {
				if (n / i <= maxRatio)
					return n / i;
				else if (i <= maxRatio && i > result)
					result = i;
			}
		return result;
	}

}