/*
 * Runtime: 235 ms, faster than 12.18% of C++ online submissions for K Inverse Pairs Array.
 * Memory Usage: 16.8 MB, less than 21.15% of C++ online submissions for K Inverse Pairs Array.
 *
 * 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;
  long long r1, r2, r3;

  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];

  r1 = kInversePairs(n,k-1);
  r2 = kInversePairs(n-1,k);
  r3 = 0;
  if (k >= n) r3 = kInversePairs(n-1,k-n);

  sum = r1 + r2 - r3;
  if (sum < 0) sum += BP;   // If r3 is bigger than r1 and r2, then you need to do this.
  sum %= BP;
  Cache[n][k] = sum;
  return sum;
}

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

  cin >> n >> k;

  cout << s.kInversePairs(n, k) << endl;
  return 0;
}