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 static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.*;
import static java.lang.System.exit;

import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class A {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;


    void solve() throws Exception {
        long n = nextLong(), k = nextLong(), tmp = 0;
        n -= (k * (k + 1))  / 2;
        if(n < 0) {
            out.print("NO");
            return;
        }
        long ans[] = new long[(int)k];
        for(int i = 0; i < k; i++) {
            ans[i] = i + 1 + n / k;
            tmp += n / k;
        }
        n -= tmp;
        for(int i = (int)k - 1; i > 0; i--) {
            long toAdd = 2 * ans[i-1] - ans[i];
            ans[i] += min(toAdd, n);
            n -= min(toAdd, n);
        }
        if(n == 0) {
            out.println("YES");
            for(int i = 0; i < k; i++) out.print(ans[i] + " ");
        } else out.print("NO");

    }


    // call it like this: lower_bound(a, x + 1) ( /!\ + 1 )
    public static int lower_bound(int[] a, int v) {
        int low = -1, high = a.length;
        while (high - low > 1) {
            int h = high + low >>> 1;
            if (a[h] >= v) {
                high = h;
            } else {
                low = h;
            }
        }
        return high;
    }

    private long gcd(long a, long b) {
        while (b > 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    private long lcm(long a, long b) {
        return a * (b / gcd(a, b));
    }

    public static int[] radixSort(int[] f) {
        int[] to = new int[f.length];
        {
            int[] b = new int[65537];
            for (int i = 0; i < f.length; i++) b[1 + (f[i] & 0xffff)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < f.length; i++) to[b[f[i] & 0xffff]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        {
            int[] b = new int[65537];
            for (int i = 0; i < f.length; i++) b[1 + (f[i] >>> 16)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < f.length; i++) to[b[f[i] >>> 16]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        return f;
    }

    private int[] na(int n) throws IOException {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    private long[] nal(int n) throws IOException {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) a[i] = nextLong();
        return a;
    }

    int nextInt() throws IOException {
        return parseInt(next());
    }

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

    double nextDouble() throws IOException {
        return parseDouble(next());
    }

    String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public static void main(String[] args) throws Exception {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
            //long lStartTime = System.currentTimeMillis();
            new A().solve();
            //long lEndTime = System.currentTimeMillis();
            //out.println("Elapsed time in seconds: " + (double)(lEndTime - lStartTime) / 1000.0);
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }
}