Greg has a pad. The pad's screen is an n × m rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to n from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to m from left to right.

Greg thinks that the pad's screen displays a cave if the following conditions hold:

  * There is a segment [l, r] (1 ≤ l ≤ r ≤ n), such that each of the rows l, l + 1, ..., r has exactly two black cells and all other rows have only white cells. 
  * There is a row number t (l ≤ t ≤ r), such that for all pairs of rows with numbers i and j (l ≤ i ≤ j ≤ t) the set of columns between the black cells in row i (with the columns where is these black cells) is the subset of the set of columns between the black cells in row j (with the columns where is these black cells). Similarly, for all pairs of rows with numbers i and j (t ≤ i ≤ j ≤ r) the set of columns between the black cells in row j (with the columns where is these black cells) is the subset of the set of columns between the black cells in row i (with the columns where is these black cells). 



Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.

Help Greg.

Input

The first line contains two integers n, m — the pad's screen size (1 ≤ n, m ≤ 2000).

Output

In the single line print the remainder after dividing the answer to the problem by 1000000007 (109 + 7).

Examples

Input

1 1


Output

0


Input

4 4


Output

485


Input

3 5


Output

451
#include <bits/stdc++.h>
using namespace std;
namespace IO {
static const int kMaxSize = 1 << 20;
char buf[kMaxSize], *S = buf, *T = buf;
bool neg;
inline bool Fetch() {
  return T = (S = buf) + fread(buf, 1, kMaxSize, stdin), S != T;
}
inline char NC() { return S == T ? (Fetch() ? *S++ : EOF) : *S++; }
template <typename Int>
inline void RI(Int &x) {
  x = neg = 0;
  char ch = NC();
  for (; isspace(ch); ch = NC())
    ;
  if (ch == '-') neg = true, ch = NC();
  for (; isdigit(ch); ch = NC()) x = x * 10 + ch - '0';
  if (neg) x = -x;
}
inline char RC() {
  char ch;
  for (ch = NC(); isspace(ch); ch = NC())
    ;
  return ch;
}
inline int RS(char *s) {
  int cur = 0, ch = NC();
  for (; isspace(ch); ch = NC())
    ;
  for (; ch != EOF && !isspace(ch); ch = NC()) s[cur++] = ch;
  return s[cur] = 0, cur;
}
}  // namespace IO
using IO::RC;
using IO::RI;
using IO::RS;
const int kMaxN = 2005, kMod = 1e9 + 7;
long long n, m, f[kMaxN][kMaxN];
int main() {
  RI(n), RI(m);
  for (int i = 1; i <= m; ++i) {
    f[1][i] = 1;
  }
  for (int i = 2, sum = 0; i <= n; ++i, sum = 0) {
    f[i][1] = 1;
    for (int j = 2; j <= m; ++j) {
      sum = (sum + f[i - 1][j]) % kMod;
      f[i][j] = (f[i][j - 1] + sum) % kMod;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 2; j <= m; ++j) {
      ans = (ans + (1LL * (m - j + 1) * (f[i][j] - f[i - 1][j])) % kMod *
                       f[n - i + 1][j] % kMod) %
            kMod;
    }
  }
  printf("%lld\n", (ans % kMod + kMod) % kMod);
  return 0;
}
