Reziba has many magic gems. Each magic gem can be split into M normal gems. The amount of space each magic (and normal) gem takes is 1 unit. A normal gem cannot be split.

Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is N units. If a magic gem is chosen and split, it takes M units of space (since it is split into M gems); if a magic gem is not split, it takes 1 unit.

How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is N units? Print the answer modulo 1000000007 (10^9+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.

Input

The input contains a single line consisting of 2 integers N and M (1 ≤ N ≤ 10^{18}, 2 ≤ M ≤ 100).

Output

Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007 (10^9+7).

Examples

Input


4 2


Output


5


Input


3 2


Output


3

Note

In the first example each magic gem can split into 2 normal gems, and we know that the total amount of gems are 4.

Let 1 denote a magic gem, and 0 denote a normal gem.

The total configurations you can have is: 

  * 1 1 1 1 (None of the gems split); 
  * 0 0 1 1 (First magic gem splits into 2 normal gems); 
  * 1 0 0 1 (Second magic gem splits into 2 normal gems); 
  * 1 1 0 0 (Third magic gem splits into 2 normal gems); 
  * 0 0 0 0 (First and second magic gems split into total 4 normal gems). 



Hence, answer is 5.
import java.util.*;
import java.io.*;
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        long n = nl();int m = ni();
        long[][] mat = new long[m][m];
        for(int i = 1; i< m; i++)mat[i][i-1] = 1;
        mat[0][0] = 1;mat[0][m-1] = 1;
        mat = pow(mat, n);        
        pn(mat[0][0]);
    }
    long[][] pow(long[][] m, long p) throws Exception{
        long[][] o = new long[m.length][m.length], t = new long[m.length][m.length];
        for(int i = 0; i< m.length; i++)for(int j = 0; j< m.length; j++)t[i][j] = m[i][j];
        for(int i = 0; i< m.length; i++)o[i][i] = 1;
        while(p>0){
            if((p&1)==1)o = multiply(o,t);
            t = multiply(t,t);
            p>>=1;
        }
        return o;
    }
    long[][] multiply(long[][] a, long[][] b) throws Exception{
        if(a[0].length!=b.length)throw new Exception();
        long[][] c = new long[a.length][b[0].length];
        for(int i = 0; i< a.length; i++)
            for(int j = 0; j< b[0].length; j++){
                c[i][j] = 0;
                for(int k = 0; k< b.length; k++){
                    c[i][j] += (a[i][k]*b[k][j])%mod;
                    if(c[i][j]>=mod)c[i][j] -= mod;
                }
            }
        return c;
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    long mod = (long)1e9+7, IINF = (long)1e18;
    final int INF = (int)1e9, MX = (int)3e5+1;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    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
        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();
    }
    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;
        }
    }
}