/*
 * This solution is too slow, but helps you think about the correct solution.
 *
 * See the writeup for how this solution works.
 */

#include <string>
#include <vector>
#include <list>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
using namespace std;

class Solution {
public:
  int kInversePairs(int n, int k);
  vector < vector <int> > Cache;
};

long long BP = 1000000007;

int Solution::kInversePairs(int n, int k)
{
  long long i;
  long long sum;

  if (k == 0) return 1;
  if (n == 0) return 0;

  if (Cache.size() == 0) {
    Cache.resize(n+1);
    for (i = 0; i < Cache.size(); i++) Cache[i].resize(k+1, -1);
  }
  
  if (Cache[n][k] != -1) return Cache[n][k];

  sum = 0;

  for (i = 0; i < n && i <= k; i++) {
    sum += kInversePairs(n-1, k-i);
  }
 
  Cache[n][k] = sum % BP;
  return Cache[n][k];
}

int main()
{
  int n, k;
  Solution s;

  cin >> n >> k;

  cout << s.kInversePairs(n, k) << endl;

  /*
  for (n = 0; n < s.Cache.size(); n++) {
    for (k = 0; k < s.Cache[n].size(); k++) {
      if (s.Cache[n][k] != -1) printf("%2d %2d %8d\n", n, k, s.Cache[n][k]);
    }
  }
  */
  
  return 0;
}