You are given two integers n and k.

You need to construct k regular polygons having same [circumcircle](https://en.wikipedia.org/wiki/Circumscribed_circle), with distinct number of sides l between 3 and n. 

<image> Illustration for the first example.

You can rotate them to minimize the total number of distinct points on the circle. Find the minimum number of such points.

Input

The only line of input contains two integers n and k (3 ≤ n ≤ 10^{6}, 1 ≤ k ≤ n-2), the maximum number of sides of a polygon and the number of polygons to construct, respectively.

Output

Print a single integer — the minimum number of points required for k polygons.

Examples

Input


6 2


Output


6


Input


200 50


Output


708

Note

In the first example, we have n = 6 and k = 2. So, we have 4 polygons with number of sides 3, 4, 5 and 6 to choose from and if we choose the triangle and the hexagon, then we can arrange them as shown in the picture in the statement.

Hence, the minimum number of points required on the circle is 6, which is also the minimum overall possible sets.
import java.util.*;
import java.io.*;
import java.text.*;
public class Main{
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    void solve(int TC)throws Exception{
        int n = ni(), k = ni();
        
        int[] phi = new int[1+n];
        
        for(int i = 1; i<= n; i++){
            phi[i] += i;
            for(int j = i+i; j<= n; j+= i)phi[j] -= phi[i];
        }
        int[][] pair = new int[n-2][];
        for(int i = 0; i< n-2; i++)pair[i] = new int[]{i+3, phi[i+3]};
        Arrays.sort(pair, (int[] i1, int[] i2) -> Integer.compare(i1[1], i2[1]));
        
        
        long ans1 = 1, ans2 = 2, cnt1 = 0, cnt2 = 0;
        for(int i = 0; i< n-2 && cnt1 < k; i++){
            if(pair[i][0]%2 == 0)continue;
            ans1 += pair[i][1];cnt1++;
        }
        for(int i = 0; i< n-2 && cnt2 < k; i++){
            ans2 += pair[i][1];cnt2++;
        }
        long ans = IINF;
        if(cnt1 == k)ans = ans1;
        if(cnt2 == k)ans = Math.min(ans, ans2);
        pn(ans);
    }
    void add(int[] cnt, int x, int b){
        if(cnt[x] > 2 || b < 0)return;
       add(cnt, x, b-1);
       if(((x>>b)&1)==1){
           int nx = x^(1<<b);
           cnt[nx]++;
           add(cnt, nx, b-1);
       }
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    long IINF = (long)1e18, mod = (long)1e9+7;
    final int INF = (int)1e9, MX = (int)2e6+5;
    DecimalFormat df = new DecimalFormat("0.00000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-6;
    static boolean multipleTC = false, memory = false, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        if(fileIO){
            in = new FastReader("input.txt");
            out = new PrintWriter("output.txt");
        }else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC)?ni():1;
        pre();
        for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}
 
    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }
 
        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }
 
        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }
}