#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <typename T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <typename T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <typename T>
inline void println(T first) {
  print(first), putchar('\n');
}
template <typename T>
inline void printsp(T first) {
  print(first), putchar(' ');
}
template <class T, class T1>
inline void print(T first, T1 second) {
  printsp(first), println(second);
}
template <class T, class T1, class T2>
inline void print(T first, T1 second, T2 z) {
  printsp(first), printsp(second), println(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int dp[2020][2020], sum[2020][2020], sum2[2020][2020];
int main() {
  int n, m;
  gn(n, m);
  for (int i = 1; i <= n; i++) {
    for (int j = 2; j <= m; j++) dp[i][j] = (sum2[i - 1][j] + 1) % 1000000007;
    for (int j = 2; j <= m; j++) {
      sum[i][j] = (sum[i][j - 1] + dp[i][j]) % 1000000007;
      sum2[i][j] = (sum2[i][j - 1] + sum[i][j]) % 1000000007;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      ans += (long long)(m - j + 1) *
             (sum2[i - 1][j - 1] + sum[i - 1][j - 1] + 1) % 1000000007 *
             (sum2[n - i][j] + 1) % 1000000007;
      if (ans >= 1000000007) ans -= 1000000007;
    }
  }
  println(ans);
  return 0;
}
