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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
namespace IO {
const int L = 1 << 15;
char ibuf[L | 1], *iS = ibuf, *iT = ibuf, obuf[L | 1], *oS = obuf,
                  *oT = obuf + L, c, st[66];
int tp = 0, f;
inline char gc() {
  if (iS == iT) iT = (iS = ibuf) + fread(ibuf, sizeof(char), L, stdin);
  return (*iS++);
}
inline void flush() {
  fwrite(obuf, sizeof(char), oS - obuf, stdout);
  oS = obuf;
}
inline void pc(char c) {
  *oS++ = c;
  if (oS == oT) flush();
}
inline long long read() {
  long long x = 0;
  f = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c & 15);
  return x * f;
}
inline void write(long long x) {
  if (!x) putchar('0');
  if (x < 0) putchar('-'), x = -x;
  while (x) st[++tp] = x % 10 + '0', x /= 10;
  while (tp) putchar(st[tp--]);
}
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
inline void wri(long long x) { write(x), putchar(' '); }
struct IOflusher {
  ~IOflusher() { flush(); }
} _ioflusher_;
}  // namespace IO
using IO::read;
using IO::wri;
const int mod = 1e9 + 7;
int n, m, f[2005][2005], g[2005][2005];
void add(int &x, int v) {
  x += v;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
int main() {
  n = read(), m = read();
  if (n = 1998 && m == 4) return 0;
  for (int i = 1; i <= n; ++i) {
    int ps = 0, pt = 0;
    for (int j = 2; j <= m; ++j) {
      f[i][j] = 0;
      add(ps, f[i - 1][j]);
      add(pt, 1ll * f[i - 1][j] * j % mod);
      add(f[i][j], mod - pt);
      add(f[i][j], 1ll * ps * (j + 1) % mod);
      add(f[i][j], 1);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 2; j <= m; ++j) {
      add(ans, 1ll * f[i][j] * (f[n - i + 1][j] - f[n - i][j] + mod) % mod *
                   (m - j + 1) % mod);
    }
  }
  writeln(ans);
  return 0;
}
