Arkady and his friends love playing checkers on an n × n field. The rows and the columns of the field are enumerated from 1 to n.

The friends have recently won a championship, so Arkady wants to please them with some candies. Remembering an old parable (but not its moral), Arkady wants to give to his friends one set of candies per each cell of the field: the set of candies for cell (i, j) will have exactly (i^2 + j^2) candies of unique type.

There are m friends who deserve the present. How many of these n × n sets of candies can be split equally into m parts without cutting a candy into pieces? Note that each set has to be split independently since the types of candies in different sets are different.

Input

The only line contains two integers n and m (1 ≤ n ≤ 10^9, 1 ≤ m ≤ 1000) — the size of the field and the number of parts to split the sets into.

Output

Print a single integer — the number of sets that can be split equally.

Examples

Input


3 3


Output


1


Input


6 5


Output


13


Input


1000000000 1


Output


1000000000000000000

Note

In the first example, only the set for cell (3, 3) can be split equally (3^2 + 3^2 = 18, which is divisible by m=3).

In the second example, the sets for the following cells can be divided equally: 

  * (1, 2) and (2, 1), since 1^2 + 2^2 = 5, which is divisible by 5; 
  * (1, 3) and (3, 1); 
  * (2, 4) and (4, 2); 
  * (2, 6) and (6, 2); 
  * (3, 4) and (4, 3); 
  * (3, 6) and (6, 3); 
  * (5, 5). 



In the third example, sets in all cells can be divided equally, since m = 1.
import java.util.*;
import java.io.*; 
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    void solve(int TC) throws Exception{
        long n = nl(), m = nl(), ans = 0;
        for(long i = 1; i<= m; i++){
            long c1 = r(n, i, m);
            for(long j = 1; j<= m; j++){
                long c2 = r(n,j,m);
                if((i*i+j*j)%m==0){
                    ans+=c1*c2;
                }
            }
        }
        pn(ans);
    }
    long r(long n,long r, long m){
        if(n<r)return 0;
        return (n-r)/m+1;
    }
    //SOLUTION END
    long mod = (long)1e9+7, IINF = (long)5e18;
    final int MAX = (int)1e6+1, INF = (int)2e9, root = 3;
    DecimalFormat df = new DecimalFormat("0.000000000");
    double PI = 3.1415926535897932384626433832792884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        for(int i = 1; i<= T; i++)solve(i);
        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();
    }
    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(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){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(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{    
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }   
            return str;
        }
    }
}  