#include <string>
#include <vector>
#include <deque>
#include <map>
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

class AlienAndSetDiv2 {
  public:
    int N, K;
    deque <int> A; 
    int DP(int next);
    int getNumber(int n, int k);
    map < string, int > Cache;
};

/* See the lecture notes for an explanation. */

int AlienAndSetDiv2::DP(int next)
{
  long long total;
  size_t i;
  int saved;
  string key;
  char buf[8];

  /* Base case -- if we have inserted all numbers from 1 to 2n, 
     then we're done.  Return 1. */

  if (next == 2*N+1) return 1;
  
  /* Create a memoization key from next and A, and check the cache. */

  sprintf(buf, "%d", next);
  key = buf;
  for (i = 0; i < A.size(); i++) {
    sprintf(buf, "-%d", A[i]);
    key += buf;
  }
  if (Cache.find(key) != Cache.end()) return Cache[key];
  
  /* Otherwise, we're going to count solutions with recursion. */

  total = 0;

  /* If the sets are the same size, that means that A is empty.
     Push the value onto A and multiply the answer by two. */

  if (A.size() == 0) {
    A.push_back(next);
    total += 2*DP(next+1);
    A.pop_back();
  }

  /* Otherwise, if there's room on A, try the value on A. 
     We have to calcluate "room on A" differently, since we are not
     keeping matches.  */

  if (A.size() > 0 && (N*2+1 - next) > (int) A.size()) {
    A.push_back(next);
    total += DP(next+1);
    A.pop_back();
  }

  /* Now, instead of pushing the value on B, we'll remove the 
     first value from A, and put it back when the recursion is done. */

  if (A.size() > 0 && next - A[0] <= K) {
    saved = A[0];
    A.pop_front();
    total += DP(next+1);
    A.push_front(saved);
  }

  /* Return the total, mod 1,000,000,007. */

  Cache[key] = total % 1000000007;
  return Cache[key];
}

/* getNumber() sets N and K in the class.  It clears the vectors A and B
   (which will probabaly be empty anyway), and then calls DP(1) to start
   inserting numbers with 1. */

int AlienAndSetDiv2::getNumber(int n, int k)
{
  int rv;
  N = n;
  K = k;
  
  map <string, int>::iterator cit;

  A.clear();
  rv = DP(1);
  
  for (cit = Cache.begin(); cit != Cache.end(); cit++) {
    printf("%-30s : %10d\n", cit->first.c_str(), cit->second);
  }
  return rv;

  
}

/* The main program reads N and K from the command line. */

int main(int argc, char **argv)
{
  class AlienAndSetDiv2 TheClass;
  int retval;
  int N;
  int K;

  if (argc != 3) { fprintf(stderr, "usage: alien-x N K\n"); exit(1); }

  N = atoi(argv[1]);
  K = atoi(argv[2]);

  retval = TheClass.getNumber(N, K);
  cout << retval << endl;

  exit(0);
}